Prove: $x^4 + y^4 + z^4 - 4(x^3 + y^3 + z^3) + 5(x^2 + y^2 + z^2) \leqslant 4 $ Prove that $$x^4 + y^4 + z^4 - 4(x^3 + y^3 + z^3) + 5(x^2 + y^2 + z^2) \leqslant 4$$ for all $x, y, z \geqslant 0$ satisfying $x + y + z = 2$. When does equality occur?

Here is my aproach: 
For each $x\in [0,2]$ we have: $$x^4-4x^3+5x^2-2x\leq 0$$
since $$(x-1)^2(x-2)x\leq 0$$
Equality holds for any permutation of $0,0,2$ or $0,1,1$.  

Any idea how to solve this with Muirhead or Am-Gm or Cauchy?

The homogenization gives
$$\frac{1}{4}(x+y+z)^4-\frac{5}{4}(x^2+y^2+z^2)(x+y+z)^2+2(x+y+z)(x^3+y^3+z^3)-x^4-y^4-z^4\geq0$$ or
$$\sum_{cyc}(x^3y+x^3z-2x^2y^2+x^2yz)\geq0,$$ which is true by AM-GM or by Muirhead because $(3,1,0)\succ(2,2,0).$
