<s>[INST] Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality:
$$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$
and I calculated partial derivatives:
$$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$
$$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$
I noticed that:
$$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$
I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).
 [/INST] (Edited version). By AM-GM we have:
$$ \sum_{cyc}(\frac{x^3}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8})$$
$$ \geq \sum_{cyc}3 \sqrt[3]{\frac{x^3}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}{8}}$$
$$=\frac{3}{4}(x+y+z)$$
we have:
$$
\sum_{cyc}(\frac{x^3}{(1+y)(1+z)}\geq \frac{3}{4}(x+y+z) - \sum_{cyc}(\frac{1+y}{8} + \frac{1+z}{8})
$$
$$
=\frac{2}{4}(x+y+z)-\frac{3}{4}\geq \frac{1}{2}\cdot 3 \sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{4}.
$$Q.E.D.
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