Prove $frac{1}{3}(x+y+z)^2 geq xy + yz + xz.$
up vote
2
down vote
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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.
inequality
add a comment |
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.
inequality
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
add a comment |
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.
inequality
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
inequality
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
add a comment |
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
add a comment |
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}
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
add a comment |
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 $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$
add a comment |
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 ${2 + 2 lambda, 2 - lambda, 2 - lambda }$
add a comment |
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 $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$
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 usege
to get $ge$. You can usesqrt{ab}
to get $sqrt{ab}$ rather thansqrt ab
, which gives $sqrt ab$. You can userRightarrow
$Rightarrow$ orimplies
$implies$ which looks a bit better than $=>$.
– Martin Sleziak
Feb 7 '16 at 7:20
add a comment |
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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.
add a comment |
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.
add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
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}
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
add a comment |
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}
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
add a comment |
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}
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}
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
add a comment |
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
add a comment |
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 $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$
add a comment |
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 $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$
add a comment |
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 $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$
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 $$(x-y)^2+(y-z)^2+(x-z)^2geq 0.$$
edited Feb 7 '16 at 4:37
answered Feb 7 '16 at 4:03
Colin Defant
677312
677312
add a comment |
add a comment |
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 ${2 + 2 lambda, 2 - lambda, 2 - lambda }$
add a comment |
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 ${2 + 2 lambda, 2 - lambda, 2 - lambda }$
add a comment |
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 ${2 + 2 lambda, 2 - lambda, 2 - lambda }$
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 ${2 + 2 lambda, 2 - lambda, 2 - lambda }$
edited Feb 7 '16 at 5:42
answered Feb 7 '16 at 4:11
Will Jagy
100k597198
100k597198
add a comment |
add a comment |
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 $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$
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 usege
to get $ge$. You can usesqrt{ab}
to get $sqrt{ab}$ rather thansqrt ab
, which gives $sqrt ab$. You can userRightarrow
$Rightarrow$ orimplies
$implies$ which looks a bit better than $=>$.
– Martin Sleziak
Feb 7 '16 at 7:20
add a comment |
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 $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$
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 usege
to get $ge$. You can usesqrt{ab}
to get $sqrt{ab}$ rather thansqrt ab
, which gives $sqrt ab$. You can userRightarrow
$Rightarrow$ orimplies
$implies$ which looks a bit better than $=>$.
– Martin Sleziak
Feb 7 '16 at 7:20
add a comment |
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 $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$
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 $$S_a^2 + S_b^2 + S_c^2 ge S_aS_b + S_bS_c + S_cS_a$$
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 usege
to get $ge$. You can usesqrt{ab}
to get $sqrt{ab}$ rather thansqrt ab
, which gives $sqrt ab$. You can userRightarrow
$Rightarrow$ orimplies
$implies$ which looks a bit better than $=>$.
– Martin Sleziak
Feb 7 '16 at 7:20
add a comment |
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 usege
to get $ge$. You can usesqrt{ab}
to get $sqrt{ab}$ rather thansqrt ab
, which gives $sqrt ab$. You can userRightarrow
$Rightarrow$ orimplies
$implies$ 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
add a comment |
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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.
add a comment |
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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.
add a comment |
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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.
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,
$2(x^2 + y^2 + z^2 - xy - yz - xz) ge 0 implies x^2 + y^2 + z^2 ge xy + yz + xz$.
answered Feb 7 '16 at 4:37
Amad27
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$frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$
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.
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$frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$
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.
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up vote
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$frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$
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.
$frac{1}{3}(x+y+z)^2 = frac{2}{3}(xy + zx + yz) + frac{1}{3}(y^2 + z^2 + x^2) geq xy + yz + xz.$
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.
edited Feb 7 '16 at 5:49
answered Feb 7 '16 at 4:41
MathematicsStudent1122
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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