Prove $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$











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Prove that for nonnegative $x,y,z$ that $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$




I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.



The reason I am asking this is because it comes from the solution below where it claims it is true.




We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a le frac{1}{3}(S_a+S_b+S_c)^2=frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.











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  • use chebyshev .
    – Shakul Pathak
    Feb 7 '16 at 4:12












  • You may notice it's true for all reals.
    – Macavity
    Feb 7 '16 at 5:56















up vote
2
down vote

favorite













Prove that for nonnegative $x,y,z$ that $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$




I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.



The reason I am asking this is because it comes from the solution below where it claims it is true.




We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a le frac{1}{3}(S_a+S_b+S_c)^2=frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.











share|cite|improve this question
























  • use chebyshev .
    – Shakul Pathak
    Feb 7 '16 at 4:12












  • You may notice it's true for all reals.
    – Macavity
    Feb 7 '16 at 5:56













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Prove that for nonnegative $x,y,z$ that $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$




I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.



The reason I am asking this is because it comes from the solution below where it claims it is true.




We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a le frac{1}{3}(S_a+S_b+S_c)^2=frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.











share|cite|improve this question
















Prove that for nonnegative $x,y,z$ that $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$




I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.



The reason I am asking this is because it comes from the solution below where it claims it is true.




We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a le frac{1}{3}(S_a+S_b+S_c)^2=frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.








inequality






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edited Nov 14 at 6:51









Robert Howard

1,612622




1,612622










asked Feb 7 '16 at 3:43









user19405892

7,54131053




7,54131053












  • use chebyshev .
    – Shakul Pathak
    Feb 7 '16 at 4:12












  • You may notice it's true for all reals.
    – Macavity
    Feb 7 '16 at 5:56


















  • use chebyshev .
    – Shakul Pathak
    Feb 7 '16 at 4:12












  • You may notice it's true for all reals.
    – Macavity
    Feb 7 '16 at 5:56
















use chebyshev .
– Shakul Pathak
Feb 7 '16 at 4:12






use chebyshev .
– Shakul Pathak
Feb 7 '16 at 4:12














You may notice it's true for all reals.
– Macavity
Feb 7 '16 at 5:56




You may notice it's true for all reals.
– Macavity
Feb 7 '16 at 5:56










6 Answers
6






active

oldest

votes

















up vote
8
down vote



accepted










Here's my attempt to expand and collect terms:
begin{align}
frac 13(x + y + z)^2 & stackrel{?}{ge} xy + xz + yz\
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & stackrel{?}{ge} 3xy + 3xz + 3yz \
x^2 + y^2 + z^2 - xy - xz - yz & stackrel{?}{ge} 0 \
frac 12(x - y)^2 + frac 12(x - z)^2 + frac 12(y - z)^2 & stackrel{?}{ge} 0.
end{align}






share|cite|improve this answer





















  • Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
    – user230452
    Feb 7 '16 at 4:25


















up vote
3
down vote













If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xzgeq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$






share|cite|improve this answer






























    up vote
    2
    down vote













    Just so you know, with real constant $lambda,$ the quadratic form
    $$ x^2 + y^2 + z^2 + lambda (yz + zx + xy) $$
    is positive definite for $-1 < lambda < 2.$ It is positive semidefinite for $lambda = 2$ and for $lambda = -1.$



    https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia



    Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
    $$
    left(
    begin{array}{ccc}
    2 & lambda & lambda \
    lambda & 2 & lambda \
    lambda & lambda & 2 \
    end{array}
    right)
    $$
    Note that the eigenvalues are ${2 + 2 lambda, 2 - lambda, 2 - lambda }$






    share|cite|improve this answer






























      up vote
      2
      down vote













      The equation, after simplification, reduces to
      $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$



      We know that A.M $ge$ G.M or $${a+bover 2} ge sqrt{ab} implies {(S_a^2 + S_b^2)over 2} ge S_aS_b$$



      Extend this to the other variables: $${(S_b^2 + S_c^2)over 2} ge S_bS_c qquad {(S_c^2 + S_a^2)over 2} ge S_cS_a$$



      Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$






      share|cite|improve this answer























      • I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
        – Martin R
        Feb 7 '16 at 4:12












      • I hope that's better @MartinR
        – Win Vineeth
        Feb 7 '16 at 4:23






      • 1




        BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
        – Martin Sleziak
        Feb 7 '16 at 7:20




















      up vote
      1
      down vote













      Like Tunococ's solution but with a modification.



      After expanding, we need to show,



      $x^2 + y^2 + z^2 ge xy + xz + yz$



      By the Cauchy-Schwarz inequality,



      $3(x^2 + y^2 + z^2) ge (x + y + z)^2$



      This means,



      $2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.






      share|cite|improve this answer




























        up vote
        1
        down vote













        $frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$



        since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.






        share|cite|improve this answer























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          6 Answers
          6






          active

          oldest

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          6 Answers
          6






          active

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          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          8
          down vote



          accepted










          Here's my attempt to expand and collect terms:
          begin{align}
          frac 13(x + y + z)^2 & stackrel{?}{ge} xy + xz + yz\
          x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & stackrel{?}{ge} 3xy + 3xz + 3yz \
          x^2 + y^2 + z^2 - xy - xz - yz & stackrel{?}{ge} 0 \
          frac 12(x - y)^2 + frac 12(x - z)^2 + frac 12(y - z)^2 & stackrel{?}{ge} 0.
          end{align}






          share|cite|improve this answer





















          • Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
            – user230452
            Feb 7 '16 at 4:25















          up vote
          8
          down vote



          accepted










          Here's my attempt to expand and collect terms:
          begin{align}
          frac 13(x + y + z)^2 & stackrel{?}{ge} xy + xz + yz\
          x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & stackrel{?}{ge} 3xy + 3xz + 3yz \
          x^2 + y^2 + z^2 - xy - xz - yz & stackrel{?}{ge} 0 \
          frac 12(x - y)^2 + frac 12(x - z)^2 + frac 12(y - z)^2 & stackrel{?}{ge} 0.
          end{align}






          share|cite|improve this answer





















          • Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
            – user230452
            Feb 7 '16 at 4:25













          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          Here's my attempt to expand and collect terms:
          begin{align}
          frac 13(x + y + z)^2 & stackrel{?}{ge} xy + xz + yz\
          x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & stackrel{?}{ge} 3xy + 3xz + 3yz \
          x^2 + y^2 + z^2 - xy - xz - yz & stackrel{?}{ge} 0 \
          frac 12(x - y)^2 + frac 12(x - z)^2 + frac 12(y - z)^2 & stackrel{?}{ge} 0.
          end{align}






          share|cite|improve this answer












          Here's my attempt to expand and collect terms:
          begin{align}
          frac 13(x + y + z)^2 & stackrel{?}{ge} xy + xz + yz\
          x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & stackrel{?}{ge} 3xy + 3xz + 3yz \
          x^2 + y^2 + z^2 - xy - xz - yz & stackrel{?}{ge} 0 \
          frac 12(x - y)^2 + frac 12(x - z)^2 + frac 12(y - z)^2 & stackrel{?}{ge} 0.
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 7 '16 at 4:09









          Tunococ

          8,1211931




          8,1211931












          • Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
            – user230452
            Feb 7 '16 at 4:25


















          • Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
            – user230452
            Feb 7 '16 at 4:25
















          Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
          – user230452
          Feb 7 '16 at 4:25




          Beautiful factorisation. The idea didn't strike me. A lesson for OP here is to solve a simpler problem when you can't solve the main problem which in this case would be (x+y)^2/3 > xy . Solve an analogous problem with two variables instead. :) . Well done.
          – user230452
          Feb 7 '16 at 4:25










          up vote
          3
          down vote













          If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xzgeq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$






          share|cite|improve this answer



























            up vote
            3
            down vote













            If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xzgeq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$






            share|cite|improve this answer

























              up vote
              3
              down vote










              up vote
              3
              down vote









              If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xzgeq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$






              share|cite|improve this answer














              If you multiply both sides by $6$ and then expand the square, you get $$2x^2+2y^2+2z^2+4xy+4yz+4xzgeq 6xy+6yz+6xz,$$ which you can rewrite as $$(x^2+y^2)+(y^2+z^2)+(x^2+z^2)geq 2xy+2yz+2xz.$$ Then rewrite this as $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 7 '16 at 4:37

























              answered Feb 7 '16 at 4:03









              Colin Defant

              677312




              677312






















                  up vote
                  2
                  down vote













                  Just so you know, with real constant $lambda,$ the quadratic form
                  $$ x^2 + y^2 + z^2 + lambda (yz + zx + xy) $$
                  is positive definite for $-1 < lambda < 2.$ It is positive semidefinite for $lambda = 2$ and for $lambda = -1.$



                  https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia



                  Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
                  $$
                  left(
                  begin{array}{ccc}
                  2 & lambda & lambda \
                  lambda & 2 & lambda \
                  lambda & lambda & 2 \
                  end{array}
                  right)
                  $$
                  Note that the eigenvalues are ${2 + 2 lambda, 2 - lambda, 2 - lambda }$






                  share|cite|improve this answer



























                    up vote
                    2
                    down vote













                    Just so you know, with real constant $lambda,$ the quadratic form
                    $$ x^2 + y^2 + z^2 + lambda (yz + zx + xy) $$
                    is positive definite for $-1 < lambda < 2.$ It is positive semidefinite for $lambda = 2$ and for $lambda = -1.$



                    https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia



                    Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
                    $$
                    left(
                    begin{array}{ccc}
                    2 & lambda & lambda \
                    lambda & 2 & lambda \
                    lambda & lambda & 2 \
                    end{array}
                    right)
                    $$
                    Note that the eigenvalues are ${2 + 2 lambda, 2 - lambda, 2 - lambda }$






                    share|cite|improve this answer

























                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      Just so you know, with real constant $lambda,$ the quadratic form
                      $$ x^2 + y^2 + z^2 + lambda (yz + zx + xy) $$
                      is positive definite for $-1 < lambda < 2.$ It is positive semidefinite for $lambda = 2$ and for $lambda = -1.$



                      https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia



                      Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
                      $$
                      left(
                      begin{array}{ccc}
                      2 & lambda & lambda \
                      lambda & 2 & lambda \
                      lambda & lambda & 2 \
                      end{array}
                      right)
                      $$
                      Note that the eigenvalues are ${2 + 2 lambda, 2 - lambda, 2 - lambda }$






                      share|cite|improve this answer














                      Just so you know, with real constant $lambda,$ the quadratic form
                      $$ x^2 + y^2 + z^2 + lambda (yz + zx + xy) $$
                      is positive definite for $-1 < lambda < 2.$ It is positive semidefinite for $lambda = 2$ and for $lambda = -1.$



                      https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia



                      Why not. Sylvester is applied to the Hessian matrix of second partial derivatives,
                      $$
                      left(
                      begin{array}{ccc}
                      2 & lambda & lambda \
                      lambda & 2 & lambda \
                      lambda & lambda & 2 \
                      end{array}
                      right)
                      $$
                      Note that the eigenvalues are ${2 + 2 lambda, 2 - lambda, 2 - lambda }$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Feb 7 '16 at 5:42

























                      answered Feb 7 '16 at 4:11









                      Will Jagy

                      100k597198




                      100k597198






















                          up vote
                          2
                          down vote













                          The equation, after simplification, reduces to
                          $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$



                          We know that A.M $ge$ G.M or $${a+bover 2} ge sqrt{ab} implies {(S_a^2 + S_b^2)over 2} ge S_aS_b$$



                          Extend this to the other variables: $${(S_b^2 + S_c^2)over 2} ge S_bS_c qquad {(S_c^2 + S_a^2)over 2} ge S_cS_a$$



                          Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$






                          share|cite|improve this answer























                          • I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                            – Martin R
                            Feb 7 '16 at 4:12












                          • I hope that's better @MartinR
                            – Win Vineeth
                            Feb 7 '16 at 4:23






                          • 1




                            BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                            – Martin Sleziak
                            Feb 7 '16 at 7:20

















                          up vote
                          2
                          down vote













                          The equation, after simplification, reduces to
                          $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$



                          We know that A.M $ge$ G.M or $${a+bover 2} ge sqrt{ab} implies {(S_a^2 + S_b^2)over 2} ge S_aS_b$$



                          Extend this to the other variables: $${(S_b^2 + S_c^2)over 2} ge S_bS_c qquad {(S_c^2 + S_a^2)over 2} ge S_cS_a$$



                          Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$






                          share|cite|improve this answer























                          • I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                            – Martin R
                            Feb 7 '16 at 4:12












                          • I hope that's better @MartinR
                            – Win Vineeth
                            Feb 7 '16 at 4:23






                          • 1




                            BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                            – Martin Sleziak
                            Feb 7 '16 at 7:20















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          The equation, after simplification, reduces to
                          $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$



                          We know that A.M $ge$ G.M or $${a+bover 2} ge sqrt{ab} implies {(S_a^2 + S_b^2)over 2} ge S_aS_b$$



                          Extend this to the other variables: $${(S_b^2 + S_c^2)over 2} ge S_bS_c qquad {(S_c^2 + S_a^2)over 2} ge S_cS_a$$



                          Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$






                          share|cite|improve this answer














                          The equation, after simplification, reduces to
                          $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$



                          We know that A.M $ge$ G.M or $${a+bover 2} ge sqrt{ab} implies {(S_a^2 + S_b^2)over 2} ge S_aS_b$$



                          Extend this to the other variables: $${(S_b^2 + S_c^2)over 2} ge S_bS_c qquad {(S_c^2 + S_a^2)over 2} ge S_cS_a$$



                          Adding all the inequalities, we get $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 14 at 5:27









                          Robert Howard

                          1,612622




                          1,612622










                          answered Feb 7 '16 at 4:01









                          Win Vineeth

                          3,275427




                          3,275427












                          • I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                            – Martin R
                            Feb 7 '16 at 4:12












                          • I hope that's better @MartinR
                            – Win Vineeth
                            Feb 7 '16 at 4:23






                          • 1




                            BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                            – Martin Sleziak
                            Feb 7 '16 at 7:20




















                          • I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                            – Martin R
                            Feb 7 '16 at 4:12












                          • I hope that's better @MartinR
                            – Win Vineeth
                            Feb 7 '16 at 4:23






                          • 1




                            BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                            – Martin Sleziak
                            Feb 7 '16 at 7:20


















                          I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                          – Martin R
                          Feb 7 '16 at 4:12






                          I find your answer difficult to read, what is "Sa^2"? This site uses TeX/MathJax for typesetting formulas, compare meta.math.stackexchange.com/questions/5020/….
                          – Martin R
                          Feb 7 '16 at 4:12














                          I hope that's better @MartinR
                          – Win Vineeth
                          Feb 7 '16 at 4:23




                          I hope that's better @MartinR
                          – Win Vineeth
                          Feb 7 '16 at 4:23




                          1




                          1




                          BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                          – Martin Sleziak
                          Feb 7 '16 at 7:20






                          BTW you can use ge to get $ge$. You can use sqrt{ab} to get $sqrt{ab}$ rather than sqrt ab, which gives $sqrt ab$. You can user Rightarrow $Rightarrow$ or implies $implies$ which looks a bit better than $=>$.
                          – Martin Sleziak
                          Feb 7 '16 at 7:20












                          up vote
                          1
                          down vote













                          Like Tunococ's solution but with a modification.



                          After expanding, we need to show,



                          $x^2 + y^2 + z^2 ge xy + xz + yz$



                          By the Cauchy-Schwarz inequality,



                          $3(x^2 + y^2 + z^2) ge (x + y + z)^2$



                          This means,



                          $2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote













                            Like Tunococ's solution but with a modification.



                            After expanding, we need to show,



                            $x^2 + y^2 + z^2 ge xy + xz + yz$



                            By the Cauchy-Schwarz inequality,



                            $3(x^2 + y^2 + z^2) ge (x + y + z)^2$



                            This means,



                            $2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.






                            share|cite|improve this answer























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Like Tunococ's solution but with a modification.



                              After expanding, we need to show,



                              $x^2 + y^2 + z^2 ge xy + xz + yz$



                              By the Cauchy-Schwarz inequality,



                              $3(x^2 + y^2 + z^2) ge (x + y + z)^2$



                              This means,



                              $2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.






                              share|cite|improve this answer












                              Like Tunococ's solution but with a modification.



                              After expanding, we need to show,



                              $x^2 + y^2 + z^2 ge xy + xz + yz$



                              By the Cauchy-Schwarz inequality,



                              $3(x^2 + y^2 + z^2) ge (x + y + z)^2$



                              This means,



                              $2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Feb 7 '16 at 4:37









                              Amad27

                              5,32321652




                              5,32321652






















                                  up vote
                                  1
                                  down vote













                                  $frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$



                                  since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.






                                  share|cite|improve this answer



























                                    up vote
                                    1
                                    down vote













                                    $frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$



                                    since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.






                                    share|cite|improve this answer

























                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      $frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$



                                      since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.






                                      share|cite|improve this answer














                                      $frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$



                                      since per Cauchy Schwarz we have $(y^2 + z^2 + x^2)^2geq (yz + xy + xz)^2$ which implies $(y^2 + z^2 + x^2) geq (yz + xy + xz)$ since square-rooting preserves order assuming a positive argument.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Feb 7 '16 at 5:49

























                                      answered Feb 7 '16 at 4:41









                                      MathematicsStudent1122

                                      7,75422159




                                      7,75422159






























                                           

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