Maximum of $ab+2bc+3ca$ with $a^4+b^4+c^4=1$












4












$begingroup$


Let $a,b,cin mathbb R^+$ with $a^4+b^4+c^4=1$. What is the maximal value $ab+2bc+3ca$ can take?

I tried using Cauchy-Schwarz several different ways and the best upper bound I got was $sqrt{14}$, but it was never sharp.

Numerical search suggests that the maximum occurs at about $a=0.763316$, $b=0.697312$, $c=0.80698$ with $ab+2bc+3ca=3.505647$, though I couldn't find any valuable relation between these numbers and the rationals.






multivariable-calculus inequality optimization lagrange-multiplier maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
    $endgroup$
    – Moo
    Nov 30 '18 at 18:26












  • $begingroup$
    But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 18:28










  • $begingroup$
    @Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
    $endgroup$
    – Richard
    Nov 30 '18 at 18:30










  • $begingroup$
    @ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
    $endgroup$
    – Moo
    Nov 30 '18 at 18:35






  • 1




    $begingroup$
    @Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 19:15
















4












$begingroup$


Let $a,b,cin mathbb R^+$ with $a^4+b^4+c^4=1$. What is the maximal value $ab+2bc+3ca$ can take?

I tried using Cauchy-Schwarz several different ways and the best upper bound I got was $sqrt{14}$, but it was never sharp.

Numerical search suggests that the maximum occurs at about $a=0.763316$, $b=0.697312$, $c=0.80698$ with $ab+2bc+3ca=3.505647$, though I couldn't find any valuable relation between these numbers and the rationals.






multivariable-calculus inequality optimization lagrange-multiplier maxima-minima







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
    $endgroup$
    – Moo
    Nov 30 '18 at 18:26












  • $begingroup$
    But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 18:28










  • $begingroup$
    @Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
    $endgroup$
    – Richard
    Nov 30 '18 at 18:30










  • $begingroup$
    @ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
    $endgroup$
    – Moo
    Nov 30 '18 at 18:35






  • 1




    $begingroup$
    @Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 19:15














4












4








4


1



$begingroup$


Let $a,b,cin mathbb R^+$ with $a^4+b^4+c^4=1$. What is the maximal value $ab+2bc+3ca$ can take?

I tried using Cauchy-Schwarz several different ways and the best upper bound I got was $sqrt{14}$, but it was never sharp.

Numerical search suggests that the maximum occurs at about $a=0.763316$, $b=0.697312$, $c=0.80698$ with $ab+2bc+3ca=3.505647$, though I couldn't find any valuable relation between these numbers and the rationals.






multivariable-calculus inequality optimization lagrange-multiplier maxima-minima







share|cite|improve this question











$endgroup$




Let $a,b,cin mathbb R^+$ with $a^4+b^4+c^4=1$. What is the maximal value $ab+2bc+3ca$ can take?

I tried using Cauchy-Schwarz several different ways and the best upper bound I got was $sqrt{14}$, but it was never sharp.

Numerical search suggests that the maximum occurs at about $a=0.763316$, $b=0.697312$, $c=0.80698$ with $ab+2bc+3ca=3.505647$, though I couldn't find any valuable relation between these numbers and the rationals.






multivariable-calculus inequality optimization lagrange-multiplier maxima-minima




multivariable-calculus inequality optimization lagrange-multiplier maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 19:21









Michael Rozenberg

97.8k1589188




97.8k1589188










asked Nov 30 '18 at 18:21









RichardRichard

3801111




3801111








  • 1




    $begingroup$
    I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
    $endgroup$
    – Moo
    Nov 30 '18 at 18:26












  • $begingroup$
    But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 18:28










  • $begingroup$
    @Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
    $endgroup$
    – Richard
    Nov 30 '18 at 18:30










  • $begingroup$
    @ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
    $endgroup$
    – Moo
    Nov 30 '18 at 18:35






  • 1




    $begingroup$
    @Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 19:15














  • 1




    $begingroup$
    I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
    $endgroup$
    – Moo
    Nov 30 '18 at 18:26












  • $begingroup$
    But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 18:28










  • $begingroup$
    @Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
    $endgroup$
    – Richard
    Nov 30 '18 at 18:30










  • $begingroup$
    @ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
    $endgroup$
    – Moo
    Nov 30 '18 at 18:35






  • 1




    $begingroup$
    @Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
    $endgroup$
    – Thomas Andrews
    Nov 30 '18 at 19:15








1




1




$begingroup$
I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
$endgroup$
– Moo
Nov 30 '18 at 18:26






$begingroup$
I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
$endgroup$
– Moo
Nov 30 '18 at 18:26














$begingroup$
But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
$endgroup$
– Thomas Andrews
Nov 30 '18 at 18:28




$begingroup$
But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$sqrt{14+pi-7log 2}.$$
$endgroup$
– Thomas Andrews
Nov 30 '18 at 18:28












$begingroup$
@Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
$endgroup$
– Richard
Nov 30 '18 at 18:30




$begingroup$
@Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c in mathbb R^+$.
$endgroup$
– Richard
Nov 30 '18 at 18:30












$begingroup$
@ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
$endgroup$
– Moo
Nov 30 '18 at 18:35




$begingroup$
@ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
$endgroup$
– Moo
Nov 30 '18 at 18:35




1




1




$begingroup$
@Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
$endgroup$
– Thomas Andrews
Nov 30 '18 at 19:15




$begingroup$
@Moo Yeah, the output above was what I got from WA when I did that. It seems unlikely the closed forms I got were the right ones.
$endgroup$
– Thomas Andrews
Nov 30 '18 at 19:15










1 Answer
1






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oldest

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3












$begingroup$

It's enough to look for non-negative variables.



Let $f(a,b,c)=ab+2bc+3ac+lambda(a^4+b^4+c^4-1)$ and $a=xb$.



Thus, in the critical point we have
$$b+2c+4lambda a^3=a+2c+4lambda b^3=2b+3a+4lambda c^3=0,$$ which gives
$$frac{b+3c}{a^3}=frac{a+2c}{b^3}=frac{2b+3a}{c^3}.$$
From the first equation we obtain:
$$c=frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
$$frac{(x^4-1)^3}{(3-2x^3)^3}left(x+frac{2(x^4-1)}{3-2x^3}right)=2+3x$$ or
$$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.






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    active

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    3












    $begingroup$

    It's enough to look for non-negative variables.



    Let $f(a,b,c)=ab+2bc+3ac+lambda(a^4+b^4+c^4-1)$ and $a=xb$.



    Thus, in the critical point we have
    $$b+2c+4lambda a^3=a+2c+4lambda b^3=2b+3a+4lambda c^3=0,$$ which gives
    $$frac{b+3c}{a^3}=frac{a+2c}{b^3}=frac{2b+3a}{c^3}.$$
    From the first equation we obtain:
    $$c=frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
    $$frac{(x^4-1)^3}{(3-2x^3)^3}left(x+frac{2(x^4-1)}{3-2x^3}right)=2+3x$$ or
    $$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
    which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      It's enough to look for non-negative variables.



      Let $f(a,b,c)=ab+2bc+3ac+lambda(a^4+b^4+c^4-1)$ and $a=xb$.



      Thus, in the critical point we have
      $$b+2c+4lambda a^3=a+2c+4lambda b^3=2b+3a+4lambda c^3=0,$$ which gives
      $$frac{b+3c}{a^3}=frac{a+2c}{b^3}=frac{2b+3a}{c^3}.$$
      From the first equation we obtain:
      $$c=frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
      $$frac{(x^4-1)^3}{(3-2x^3)^3}left(x+frac{2(x^4-1)}{3-2x^3}right)=2+3x$$ or
      $$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
      which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        It's enough to look for non-negative variables.



        Let $f(a,b,c)=ab+2bc+3ac+lambda(a^4+b^4+c^4-1)$ and $a=xb$.



        Thus, in the critical point we have
        $$b+2c+4lambda a^3=a+2c+4lambda b^3=2b+3a+4lambda c^3=0,$$ which gives
        $$frac{b+3c}{a^3}=frac{a+2c}{b^3}=frac{2b+3a}{c^3}.$$
        From the first equation we obtain:
        $$c=frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
        $$frac{(x^4-1)^3}{(3-2x^3)^3}left(x+frac{2(x^4-1)}{3-2x^3}right)=2+3x$$ or
        $$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
        which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.






        share|cite|improve this answer











        $endgroup$



        It's enough to look for non-negative variables.



        Let $f(a,b,c)=ab+2bc+3ac+lambda(a^4+b^4+c^4-1)$ and $a=xb$.



        Thus, in the critical point we have
        $$b+2c+4lambda a^3=a+2c+4lambda b^3=2b+3a+4lambda c^3=0,$$ which gives
        $$frac{b+3c}{a^3}=frac{a+2c}{b^3}=frac{2b+3a}{c^3}.$$
        From the first equation we obtain:
        $$c=frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
        $$frac{(x^4-1)^3}{(3-2x^3)^3}left(x+frac{2(x^4-1)}{3-2x^3}right)=2+3x$$ or
        $$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
        which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 19:24

























        answered Nov 30 '18 at 19:14









        Michael RozenbergMichael Rozenberg

        97.8k1589188




        97.8k1589188






























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