Cauchy Schwartz Inequality Question: $(a^2+b^2)^3=c^2+d^2 implies frac{a^3}{c}+frac{b^3}{d}geq 1$












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If $(a^2+b^2)^3=c^2+d^2$, prove that $frac{a^3}{c}+frac{b^3}{d}geq 1$.




Please help.






inequality cauchy-schwarz-inequality holder-inequality







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  • $begingroup$
    Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 5:11
















0












$begingroup$



If $(a^2+b^2)^3=c^2+d^2$, prove that $frac{a^3}{c}+frac{b^3}{d}geq 1$.




Please help.






inequality cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 5:11














0












0








0


2



$begingroup$



If $(a^2+b^2)^3=c^2+d^2$, prove that $frac{a^3}{c}+frac{b^3}{d}geq 1$.




Please help.






inequality cauchy-schwarz-inequality holder-inequality







share|cite|improve this question











$endgroup$





If $(a^2+b^2)^3=c^2+d^2$, prove that $frac{a^3}{c}+frac{b^3}{d}geq 1$.




Please help.






inequality cauchy-schwarz-inequality holder-inequality




inequality cauchy-schwarz-inequality holder-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 6:40









Michael Rozenberg

105k1892198




105k1892198










asked Jan 18 at 5:07









SaeeSaee

387




387












  • $begingroup$
    Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 5:11


















  • $begingroup$
    Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 5:11
















$begingroup$
Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 5:11




$begingroup$
Can you explain why you think Cauchy Schwarz Inequality is the best approach for this problem
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 5:11










3 Answers
3






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oldest

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1












$begingroup$

Hint: Use
$$
(a^{2} + b^{2})(c^{2} + d^{2}) geq (ac + bd)^{2}
$$
and
$$
left( frac{a^{3}}{c} + frac{b^{3}}{d}right)(ac+bd) geq (a^{2} + b^{2})^{2}.
$$






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$endgroup$





















    1












    $begingroup$

    Because by Holder
    $$left(frac{a^3}{c}+frac{b^3}{d}right)^2(c^2+d^2)geq(a^2+b^2)^3.$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $LHS = dfrac{(a^2)^2}{ac}+ dfrac{(b^2)^2}{bd}ge dfrac{(a^2+b^2)^2}{ac+bd}= dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}ge dfrac{c^2+d^2}{(a^2+b^2)cdot sqrt{a^2+b^2}cdot sqrt{c^2+d^2}}= sqrt{dfrac{c^2+d^2}{(a^2+b^2)^3}}= sqrt{1} = 1= RHS. $






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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Hint: Use
        $$
        (a^{2} + b^{2})(c^{2} + d^{2}) geq (ac + bd)^{2}
        $$
        and
        $$
        left( frac{a^{3}}{c} + frac{b^{3}}{d}right)(ac+bd) geq (a^{2} + b^{2})^{2}.
        $$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Hint: Use
          $$
          (a^{2} + b^{2})(c^{2} + d^{2}) geq (ac + bd)^{2}
          $$
          and
          $$
          left( frac{a^{3}}{c} + frac{b^{3}}{d}right)(ac+bd) geq (a^{2} + b^{2})^{2}.
          $$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint: Use
            $$
            (a^{2} + b^{2})(c^{2} + d^{2}) geq (ac + bd)^{2}
            $$
            and
            $$
            left( frac{a^{3}}{c} + frac{b^{3}}{d}right)(ac+bd) geq (a^{2} + b^{2})^{2}.
            $$






            share|cite|improve this answer









            $endgroup$



            Hint: Use
            $$
            (a^{2} + b^{2})(c^{2} + d^{2}) geq (ac + bd)^{2}
            $$
            and
            $$
            left( frac{a^{3}}{c} + frac{b^{3}}{d}right)(ac+bd) geq (a^{2} + b^{2})^{2}.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 18 at 6:21









            Seewoo LeeSeewoo Lee

            6,959927




            6,959927























                1












                $begingroup$

                Because by Holder
                $$left(frac{a^3}{c}+frac{b^3}{d}right)^2(c^2+d^2)geq(a^2+b^2)^3.$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Because by Holder
                  $$left(frac{a^3}{c}+frac{b^3}{d}right)^2(c^2+d^2)geq(a^2+b^2)^3.$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Because by Holder
                    $$left(frac{a^3}{c}+frac{b^3}{d}right)^2(c^2+d^2)geq(a^2+b^2)^3.$$






                    share|cite|improve this answer









                    $endgroup$



                    Because by Holder
                    $$left(frac{a^3}{c}+frac{b^3}{d}right)^2(c^2+d^2)geq(a^2+b^2)^3.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 18 at 6:35









                    Michael RozenbergMichael Rozenberg

                    105k1892198




                    105k1892198























                        1












                        $begingroup$

                        Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $LHS = dfrac{(a^2)^2}{ac}+ dfrac{(b^2)^2}{bd}ge dfrac{(a^2+b^2)^2}{ac+bd}= dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}ge dfrac{c^2+d^2}{(a^2+b^2)cdot sqrt{a^2+b^2}cdot sqrt{c^2+d^2}}= sqrt{dfrac{c^2+d^2}{(a^2+b^2)^3}}= sqrt{1} = 1= RHS. $






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $LHS = dfrac{(a^2)^2}{ac}+ dfrac{(b^2)^2}{bd}ge dfrac{(a^2+b^2)^2}{ac+bd}= dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}ge dfrac{c^2+d^2}{(a^2+b^2)cdot sqrt{a^2+b^2}cdot sqrt{c^2+d^2}}= sqrt{dfrac{c^2+d^2}{(a^2+b^2)^3}}= sqrt{1} = 1= RHS. $






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $LHS = dfrac{(a^2)^2}{ac}+ dfrac{(b^2)^2}{bd}ge dfrac{(a^2+b^2)^2}{ac+bd}= dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}ge dfrac{c^2+d^2}{(a^2+b^2)cdot sqrt{a^2+b^2}cdot sqrt{c^2+d^2}}= sqrt{dfrac{c^2+d^2}{(a^2+b^2)^3}}= sqrt{1} = 1= RHS. $






                            share|cite|improve this answer









                            $endgroup$



                            Using the standard and the Angel forms of the Cauchy-Schwarz inequality: $LHS = dfrac{(a^2)^2}{ac}+ dfrac{(b^2)^2}{bd}ge dfrac{(a^2+b^2)^2}{ac+bd}= dfrac{(a^2+b^2)^3}{(a^2+b^2)(ac+bd)}= dfrac{c^2+d^2}{(a^2+b^2)(ac+bd)}ge dfrac{c^2+d^2}{(a^2+b^2)cdot sqrt{a^2+b^2}cdot sqrt{c^2+d^2}}= sqrt{dfrac{c^2+d^2}{(a^2+b^2)^3}}= sqrt{1} = 1= RHS. $







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 18 at 6:42









                            DeepSeaDeepSea

                            71.3k54487




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