nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$10(81u^4-108u^2v^2+18v^4+12uw^3)+21w^4-17(27u^3-27uv^2+3w^3)\geq0$$ 
or $f(u)\geq0$, where
$$f(u)=270u^4-360u^2v^2+60v^4+40uw^3-153u^3w+153uv^2w-10w^4.$$
But by Schur $w^3\geq4uv^2-3u^3$.
Thus,
$$f'(u)=1080u^3-720uv^2+40w^3-459u^2w+153v^2w\geq$$
$$\geq621u^3-720uv^2+99w^3=9(69u^3-80uv^2+11w^3)\geq$$
$$\geq9(69u^3-80uv^2+44uv^2-33u^3)=324u(u^2-v^2)\geq0,$$
which says that $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $u$, 
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
Id est, we need to prove that:
$$(a-1)^2(20a^{10}+6a^9-8a^8-22a^7-15a^6-8a^5-a^4+6a^3+13a^2+20a+1)\geq0,$$
which is true.
Done!
