Easy $uvw$.

arqady wrote: Easy $uvw$. i don't think that official solution is with $uvw$ i am pretty sure that there is a simple nice proof

The hard part is to figure out and equality case, which is $(1,4,4)$. Then if we denote $x=a^3,y=4b^3,z=4c^3$ a full expansion and a smooth combination of AM-GM for 90 terms will end up the problem.(I hope I'm right)

GGPiku wrote: The hard part is to figure out and equality case, which is $(1,4,4)$. Then if we denote $x=a^3,y=4b^3,z=4c^3$ a full expansion and a smooth combination of AM-GM for 90 terms will end up the problem.(I hope I'm right) This will never ever work and the reason is simple: There are two more equality cases: $(4,1,4)$ and $(4,4,1)$. So you won't find an AM-GM estimate in which $x$ plays a special role. In fact, one of the official solutions argues indeed essentially with uvw. My favourite solution (which I found during that competition) is a mixing-variables approach: W.l.o.g. let $x \ge y \ge z$. Let $LHS-RHS=f(x,y,z)$. Then it is not hard to show that $f(x,y,z) \ge f(\sqrt{xy},\sqrt{xy},z)$ and from here by homogenity the problem reduces to a one-variable inequality which is then easily established.

As Tintarn said $x\ge y\ge z$. $f(x,y,z) \ge f(\sqrt{xy},\sqrt{xy},z)\iff (\sqrt{x}-\sqrt{y})^2\left[(2x+2y-9\sqrt{xy})z^2+xy\sqrt{xy}+z\cdot \sqrt{xy}(\sqrt{x}+\sqrt{y})^2\right]\ge 0$ We have to prove $$(2x+2y-9\sqrt{xy})z^2+xy\sqrt{xy}+z\cdot \sqrt{xy}(\sqrt{x}+\sqrt{y})^2\ge 0$$If $2x+2y-9\sqrt{xy}>0$ then obviously holds. If $2x+2y-9\sqrt{xy}\le 0$ it's sufficient to prove for $z=0, z=y$. First case is clear and second is: $$2y^2(x-\sqrt{xy})+y\sqrt{y}(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}-y)\ge 0$$and it's true ($x\ge y$).

Tintarn wrote: Prove that for all $x,y,z>0$, the inequality \[\frac{x+y+z}{3}+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \ge 5 \sqrt[3]{\frac{xyz}{16}}\]holds. Determine if equality can hold and if so, in which cases it occurs. WLOG, we can assume $xyz=2$. We need to prove: \[\frac{x+y+z}{3}+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \geq \frac{5}{2}\]The inequality is linear on $u$. According to uvw theorem, we only need to check $y=x$ and $z = \frac{2}{x^2}$. In this case, we get: $$\frac{(x - 2)^2 (4 x^4 + x^3 - 12 x^2 + 4 x + 4)}{6 x^2 (x^3 + 4)} \geq 0$$which is obviously true. Equality at $(x,y,z) = \left(2,2,\frac{1}{2}\right)$ and their permutations.