<s>[INST] Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't needs the condition anymore to prove this, but I couldn't get any further.
 [/INST] By AM-GM:
$$\begin{aligned}a^4+a^4+a^4+a^4+a^4+a^2+a^2+1&\ge 8\sqrt[8]{a^{4\times 5}a^{2\times 2}}\\&=8\sqrt[8]{a^{24}}\\&=8|a^3|\ge8a^3.\end{aligned}$$
Similarly for $b$ and $c$ and add them up.
</s>