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.

Hint:
It is enough to show $f(x)=5x^4+3-8x^3+2(x^2-1)=(x-1)^2(5x^2+2x+1)\ge0$ which is obvious.
