How to solve this problem using Cauchy-Schwarz inequality












2












$begingroup$


How to solve this problem? This is one of the problems in my semester examination:




For $x,yinmathbb{R}^+,x^2+y^2=1$, find the maximum value of $M=sqrt{x}+sqrt{2y}$.




I know it can be solved using the Cauchy-Schwarz inequality, but I can't solve it despite much effort with this idea.



Any help will be much appreciated! Although the examination is now over, but a solution for this problem will be very useful to me.



Thank you in advance.






inequality cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    How to solve this problem? This is one of the problems in my semester examination:




    For $x,yinmathbb{R}^+,x^2+y^2=1$, find the maximum value of $M=sqrt{x}+sqrt{2y}$.




    I know it can be solved using the Cauchy-Schwarz inequality, but I can't solve it despite much effort with this idea.



    Any help will be much appreciated! Although the examination is now over, but a solution for this problem will be very useful to me.



    Thank you in advance.






    inequality cauchy-schwarz-inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      How to solve this problem? This is one of the problems in my semester examination:




      For $x,yinmathbb{R}^+,x^2+y^2=1$, find the maximum value of $M=sqrt{x}+sqrt{2y}$.




      I know it can be solved using the Cauchy-Schwarz inequality, but I can't solve it despite much effort with this idea.



      Any help will be much appreciated! Although the examination is now over, but a solution for this problem will be very useful to me.



      Thank you in advance.






      inequality cauchy-schwarz-inequality







      share|cite|improve this question











      $endgroup$




      How to solve this problem? This is one of the problems in my semester examination:




      For $x,yinmathbb{R}^+,x^2+y^2=1$, find the maximum value of $M=sqrt{x}+sqrt{2y}$.




      I know it can be solved using the Cauchy-Schwarz inequality, but I can't solve it despite much effort with this idea.



      Any help will be much appreciated! Although the examination is now over, but a solution for this problem will be very useful to me.



      Thank you in advance.






      inequality cauchy-schwarz-inequality




      inequality cauchy-schwarz-inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 4:27







      Dũng Vũ

















      asked Dec 13 '18 at 4:20









      Dũng VũDũng Vũ

      1156




      1156






















          1 Answer
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          2












          $begingroup$

          Consider Hölder's inequality:
          $$(x^2+y^2)(1+2^{2/3})^3 geqslant (sqrt{x} + sqrt{2y})^4$$



          This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $sqrt x + sqrt{2y} $ as $(1+2^{2/3})^{3/4} approx 2.04$.




          P.S. Adding the CS inequality version as well:
          $$(x^2+y^2)(1+k) geqslant (x+sqrt k, y)^2 tag{$CS_1$}$$
          $$(x+sqrt k, y)(1+k) geqslant (sqrt x+k^{3/4}sqrt{y})^2 tag{$CS_2$}$$



          Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives
          $$(x^2+y^2)(1+k)^3 geqslant (sqrt x+k^{3/4}sqrt{y})^4$$
          Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:sqrt k$, which is possible for any $k> 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 4:50










          • $begingroup$
            I have added the two CS equivalent also ...
            $endgroup$
            – Macavity
            Dec 13 '18 at 5:05










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 5:32











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Consider Hölder's inequality:
          $$(x^2+y^2)(1+2^{2/3})^3 geqslant (sqrt{x} + sqrt{2y})^4$$



          This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $sqrt x + sqrt{2y} $ as $(1+2^{2/3})^{3/4} approx 2.04$.




          P.S. Adding the CS inequality version as well:
          $$(x^2+y^2)(1+k) geqslant (x+sqrt k, y)^2 tag{$CS_1$}$$
          $$(x+sqrt k, y)(1+k) geqslant (sqrt x+k^{3/4}sqrt{y})^2 tag{$CS_2$}$$



          Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives
          $$(x^2+y^2)(1+k)^3 geqslant (sqrt x+k^{3/4}sqrt{y})^4$$
          Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:sqrt k$, which is possible for any $k> 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 4:50










          • $begingroup$
            I have added the two CS equivalent also ...
            $endgroup$
            – Macavity
            Dec 13 '18 at 5:05










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 5:32
















          2












          $begingroup$

          Consider Hölder's inequality:
          $$(x^2+y^2)(1+2^{2/3})^3 geqslant (sqrt{x} + sqrt{2y})^4$$



          This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $sqrt x + sqrt{2y} $ as $(1+2^{2/3})^{3/4} approx 2.04$.




          P.S. Adding the CS inequality version as well:
          $$(x^2+y^2)(1+k) geqslant (x+sqrt k, y)^2 tag{$CS_1$}$$
          $$(x+sqrt k, y)(1+k) geqslant (sqrt x+k^{3/4}sqrt{y})^2 tag{$CS_2$}$$



          Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives
          $$(x^2+y^2)(1+k)^3 geqslant (sqrt x+k^{3/4}sqrt{y})^4$$
          Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:sqrt k$, which is possible for any $k> 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 4:50










          • $begingroup$
            I have added the two CS equivalent also ...
            $endgroup$
            – Macavity
            Dec 13 '18 at 5:05










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 5:32














          2












          2








          2





          $begingroup$

          Consider Hölder's inequality:
          $$(x^2+y^2)(1+2^{2/3})^3 geqslant (sqrt{x} + sqrt{2y})^4$$



          This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $sqrt x + sqrt{2y} $ as $(1+2^{2/3})^{3/4} approx 2.04$.




          P.S. Adding the CS inequality version as well:
          $$(x^2+y^2)(1+k) geqslant (x+sqrt k, y)^2 tag{$CS_1$}$$
          $$(x+sqrt k, y)(1+k) geqslant (sqrt x+k^{3/4}sqrt{y})^2 tag{$CS_2$}$$



          Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives
          $$(x^2+y^2)(1+k)^3 geqslant (sqrt x+k^{3/4}sqrt{y})^4$$
          Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:sqrt k$, which is possible for any $k> 0$.






          share|cite|improve this answer











          $endgroup$



          Consider Hölder's inequality:
          $$(x^2+y^2)(1+2^{2/3})^3 geqslant (sqrt{x} + sqrt{2y})^4$$



          This can get equality when $x:y = 1:2^{1/3}$ which is certainly possible, so this determines the maximum of $sqrt x + sqrt{2y} $ as $(1+2^{2/3})^{3/4} approx 2.04$.




          P.S. Adding the CS inequality version as well:
          $$(x^2+y^2)(1+k) geqslant (x+sqrt k, y)^2 tag{$CS_1$}$$
          $$(x+sqrt k, y)(1+k) geqslant (sqrt x+k^{3/4}sqrt{y})^2 tag{$CS_2$}$$



          Now multiplying inequalities $(CS_1)$ and $(CS_2)^2$ together gives
          $$(x^2+y^2)(1+k)^3 geqslant (sqrt x+k^{3/4}sqrt{y})^4$$
          Setting $k=2^{2/3}$ makes the RHS what we want, which is the same as the Hölder's inequality above. You can work out that equality in both CS inequalities requires $x:y=1:sqrt k$, which is possible for any $k> 0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 5:10

























          answered Dec 13 '18 at 4:42









          MacavityMacavity

          35.3k52554




          35.3k52554












          • $begingroup$
            Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 4:50










          • $begingroup$
            I have added the two CS equivalent also ...
            $endgroup$
            – Macavity
            Dec 13 '18 at 5:05










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 5:32


















          • $begingroup$
            Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 4:50










          • $begingroup$
            I have added the two CS equivalent also ...
            $endgroup$
            – Macavity
            Dec 13 '18 at 5:05










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Dũng Vũ
            Dec 13 '18 at 5:32
















          $begingroup$
          Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
          $endgroup$
          – Dũng Vũ
          Dec 13 '18 at 4:50




          $begingroup$
          Thank you very much! It is very short! However, as Hölder's inequality is a bit unfamiliar to us, could you please expand your answer and add a solution using CS inequality twice? I am looking forward to that!
          $endgroup$
          – Dũng Vũ
          Dec 13 '18 at 4:50












          $begingroup$
          I have added the two CS equivalent also ...
          $endgroup$
          – Macavity
          Dec 13 '18 at 5:05




          $begingroup$
          I have added the two CS equivalent also ...
          $endgroup$
          – Macavity
          Dec 13 '18 at 5:05












          $begingroup$
          Thank you very much!
          $endgroup$
          – Dũng Vũ
          Dec 13 '18 at 5:32




          $begingroup$
          Thank you very much!
          $endgroup$
          – Dũng Vũ
          Dec 13 '18 at 5:32


















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