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Let $a,b,c\in \mathbb R^+$ 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.

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    $\begingroup$ I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry). $\endgroup$
    – Moo
    Nov 30, 2018 at 18:26
  • $\begingroup$ But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$\sqrt{14+\pi-7\log 2}.$$ $\endgroup$ Nov 30, 2018 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 $a,b,c \in \mathbb R^+$. $\endgroup$
    – Cecilia
    Nov 30, 2018 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, 2018 at 18:35
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    $\begingroup$ @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. $\endgroup$ Nov 30, 2018 at 19:15

1 Answer 1

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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 $$b+2c+4\lambda a^3=a+2c+4\lambda b^3=2b+3a+4\lambda c^3=0,$$ 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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