Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $sum_{text{cyc}}frac{1}{a^3(b+c)}geq...












3












$begingroup$


I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:




Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$




First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
&geq 3fleft(frac{x+y+z}{3} right) \
&= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
&= frac12 (x+y+z) geq frac32end{alignat*}$$
The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)






proof-verification inequality contest-math alternative-proof jensen-inequality







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    3












    $begingroup$


    I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:




    Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$




    First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
    &geq 3fleft(frac{x+y+z}{3} right) \
    &= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
    &= frac12 (x+y+z) geq frac32end{alignat*}$$
    The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)






    proof-verification inequality contest-math alternative-proof jensen-inequality







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:




      Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$




      First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
      &geq 3fleft(frac{x+y+z}{3} right) \
      &= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
      &= frac12 (x+y+z) geq frac32end{alignat*}$$
      The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)






      proof-verification inequality contest-math alternative-proof jensen-inequality







      share|cite|improve this question











      $endgroup$




      I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:




      Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$frac{1}{a^3(b+c)} + frac{1}{b^3(a+c)} + frac{1}{c^3(a+b)}geq frac32$$




      First, substitute $frac1a = x, frac1b = y, frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$frac{x^2}{z+y} + frac{y^2}{x+z} + frac{z^2}{x+y} geq frac32$$ Now, let $f(x)=dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,zin mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$begin{alignat*}{2}frac{x^2}{y+z} + frac{y^2}{x+z} + frac{z^2}{x+y} & =frac{x^2}{S-x} + frac{y^2}{S-y} + frac{z^2}{S-z} \
      &geq 3fleft(frac{x+y+z}{3} right) \
      &= 3frac{left(frac{x+y+z}{3}right)^2}{S-frac{x+y+z}{3}} = frac{1}{3}frac{(x+y+z)^2}{frac{2x+2y+2z}{3}} = frac13 frac{(x+y+z)^2}{frac23 (x+y+z)} \
      &= frac12 (x+y+z) geq frac32end{alignat*}$$
      The final inequality follows by AM-GM ($x+y+zgeq 3sqrt[3]{xyz}=3$)






      proof-verification inequality contest-math alternative-proof jensen-inequality




      proof-verification inequality contest-math alternative-proof jensen-inequality






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      edited Dec 16 '18 at 20:07









      Blue

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      asked Dec 16 '18 at 19:49









      MarkusMarkus

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          3 Answers
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          $begingroup$

          Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              I think it's better to end your proof by C-S and AM-GM:
              $$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$






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                3 Answers
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                3 Answers
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                $begingroup$

                Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.






                    share|cite|improve this answer









                    $endgroup$



                    Your proof is nearly correct: $$f''(x)=frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 16 '18 at 19:59









                    Carl SchildkrautCarl Schildkraut

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                    11.7k11443























                        1












                        $begingroup$

                        You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.






                            share|cite|improve this answer









                            $endgroup$



                            You make a little mistake i think: $$f''(x)=frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 16 '18 at 20:02









                            Maria MazurMaria Mazur

                            46.9k1260120




                            46.9k1260120























                                1












                                $begingroup$

                                I think it's better to end your proof by C-S and AM-GM:
                                $$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  I think it's better to end your proof by C-S and AM-GM:
                                  $$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    I think it's better to end your proof by C-S and AM-GM:
                                    $$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    I think it's better to end your proof by C-S and AM-GM:
                                    $$sum_{cyc}frac{x^2}{y+z}geqfrac{(x+y+z)^2}{sumlimits_{cyc}(y+z)}=frac{1}{2}(x+y+z)geqfrac{3}{2}.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 16 '18 at 22:02









                                    Michael RozenbergMichael Rozenberg

                                    108k1895200




                                    108k1895200






























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