<s>[INST] How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
I think this is likely to turn out to be proved by Hölder, but I can't see how. Any hints will be appreciated.
 [/INST] W.L.O.G., let $a\ge b\ge c$, then by the rearrangement inequality, we have
\begin{align*}
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{b^6}{b^2+c^2}+\frac{c^6}{a^2+c^2}+\frac{a^6}{a^2+b^2},\tag{1}\\
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{c^6}{b^2+c^2}+\frac{a^6}{a^2+c^2}+\frac{b^6}{a^2+b^2}.\tag{2}
\end{align*}
Thus combine $(1)$ and $(2)$,
\begin{align*}
2&\left(\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\right)\\
&=\frac{b^6+c^6}{b^2+c^2}+\frac{a^6+c^6}{a^2+c^2}+\frac{a^6+b^6}{a^2+b^2}\\
&=(b^4-b^2c^2+c^4)+(a^4-a^2c^2+c^4)+(a^4-a^2b^2+b^4)\\
&=(a^4+b^4+c^4)+(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2)\\
&\ge a^4+b^4+c^4\tag{3}\\
&=\frac{a^4+b^4}{2}+\frac{b^4+c^4}{2}+\frac{c^4+a^4}{2}\\
&\ge \frac{a^4+b^2c^2}{2}+\frac{b^4+a^2c^2}{2}+\frac{c^4+a^2b^2}{2}\tag{4}\\
&\ge a^2bc+b^2ac+c^2ab\\
&=abc(a+b+c),
\end{align*}
where $(3)$ and $(4)$ are also used the rearrangement inequality. So the result follows.
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