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).

We need to prove that:
$$\sum_{cyc}(4x^4+4x^3)\geq\sum_{cyc}(2+3x+3xy),$$ which is true by Muirhead.
Indeed, we need to prove that:
$$\sum_{cyc}(3x^4+3x^3+x^4+x^3)\geq\sum_{cyc}(3xy+3x+1+1).$$
Now, $$\sum_{cyc}x^4\geq\sum_{cyc}xy$$ it's
$$\sum_{cyc}x^4\geq\sum_{cyc}xy\sqrt[3]{x^2y^2z^2}$$ or
$$\sum_{cyc}x^4\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{5}{3}}z^{\frac{2}{3}},$$ which is true by Muirhead because $$\left(4,0,0\right)\succ\left(\frac{5}{3},\frac{5}{3},\frac{2}{3}\right).$$
Also, $$\sum_{cyc}x^3\geq\sum_{cyc}x$$ it's
$$\sum_{cyc}x^3\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}},$$
which is true by Muirhead again  because
$$(3,0,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right).$$
Also, by AM-GM $$\sum_{cyc}x^4\geq3\sqrt[3]{x^4y^4z^4}=3=\sum_{cyc}1$$ and
$$\sum_{cyc}x^3\geq3xyz=3=\sum_{cyc}1.$$ 
