<s>[INST] Find the maximum value of:$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$ Let $a,b,c \in R$, find the maximum value of:$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$$
I have the solution:
$$5a^2+2b^2+c^2-2(ab+bc+ca)=5(a-\frac{b+c}{5})^2+\frac{1}{5}(3b-2c)^2\geqslant 0$$
which is equivalent to
$$ a^2+b^2+c^2-ab-bc-ca\geqslant \frac{c^2-3a^2}{2}$$
or
$$ 2\sqrt{\frac{a^2+b^2+c^2-ab-bc-ca}{3}}\geqslant 2\sqrt{\frac{c^2-3a^2}{6}}$$
Now, let $x=\sqrt[6]{\dfrac{c^2-3a^2}{6}}$ and the problem is find the maximum value of $P=3t^2-2t^3$, which is easy.
My question:

*

*Is there any other solution to this problem?

*I try to test it on Wolfram Alpha, the result is that the maximum value does not exist. Why ? Is Wolfram Alpha's algorithm wrong or does it use a special calculation method?

 [/INST] Alternate approach:
$a^2+b^2+c^2-ab-bc-ca$ takes its max value when $b =\frac{a+c}2$. So $a^2+b^2+c^2-ab-bc-ca=\frac 34(a-c)^2$, so we aim to maximize
$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}- |c-a|$$
Let $c-a=t$, so $c^2-3a^2=(a+t)^2-3a^2=-2a^2+2at+t^2$. This takes maximum if $a=t/2$, and $c^2-3a^2=3/2 t^2$. Therefore, we aim to maximize
$$3\sqrt[3]{\dfrac{t^2}{4}}-|t|$$
And this takes the maximum when $t=2$, and the maximum value is $1$.
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