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

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.

• I also tried on Wolfram Alpha, but it basically validates your results (there appears to be a symmetry).
– Moo
Nov 30, 2018 at 18:26
• But Wolfram gives nothing useful for a closed form for $3.505647.$ Things like $$\sqrt{14+\pi-7\log 2}.$$ Nov 30, 2018 at 18:28
• @Moo Yes, if you negate a, b and c, neither the fourth powers nor the products change. That's why I specified $a,b,c \in \mathbb R^+$. Nov 30, 2018 at 18:30
• @ThomasAndrews: Even though it does not find a closed form, you can do things like Wolfram Alpha to get closed form expressions.
– Moo
Nov 30, 2018 at 18:35
• @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. Nov 30, 2018 at 19:15

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.