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.

You can prove it as follows:
The inequality is equivalent to:
$$
\sum_{cyc}5a^4-8a^3+3a^2≥0\iff\sum_{cyc}5a^2\left(\left(a-\frac{4}{5}\right)^2-\frac{1}{25}\right)≥0\iff\sum_{cyc}a^2\left(\left(5a-4\right)^2-1\right)≥0\iff\\
\sum_{cyc}a^2(5a-4)^2≥3\iff\sqrt{\frac{1}{3}\sum_{cyc}a^2(5a-4)^2}≥1
$$
Now, by AM-QM:
$$
\sqrt{\frac{1}{3}\sum_{cyc}a^2(5a-4)^2}≥\frac{1}{3}\sum_{cyc}a(5a-4)=5-4\frac{a+b+c}{3}≥5-4\sqrt{\frac{a^2+b^2+c^2}{3}}=1
$$
And we're done.
