<s>[INST] Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$ Let $x,y,z>0$. Prove that
$$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$$
On the left-hand side we have arithmetic and harmonic means, while on the right geometric mean. The AM-GM-HM inequality states that AM $\geq$ GM $\geq$ HM, but here HM is on the wrong side of the inequality.
 [/INST] Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(u)\geq0,$ where
$$f(u)=u+\frac{w^3}{v^2}-\frac{5w}{\sqrt[3]{16}}.$$
Id est, it's enough to prove our inequality for the minimal value of $u$, which happens for equality case of two variables.
Since our inequality is symmetric and homogeneous, we can assume $y=z=1$ and $z=16a^3$
and we need to prove that:
$$\frac{16a^3+2}{3}+\frac{48a^3}{32a^3+1}\geq5a$$ or
$$(4a-1)^2(32a^4+16a^3-24a^2+a+2)\geq0$$ or
$$(4a-1)^2(2(4a^2-1)^2+a(4a-1)^2)\geq0,$$ which is obvious.
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