Interpreting an olympiad inequality as a convex function We need to prove the following inequality:

$$\frac{y^3xz}{x^3(xy+z^2)}+\frac{z^3xy}{y^3(x^2+yz)}+\frac{x^3yz}{z^3(xz+y^2)}\geq \frac{3}{2}$$

This equation is convex in each of the variables $x,y,z$. Moreover, its minimum seems to be at the origin (the equation is homogeneous, and there are not displacements of the vertex). 
Can I then say that the minima is achieved when $x=y=z$? Clearly, when $x=y=z$, the value of the inequality is $\frac{3}{2}$

$$\Longleftrightarrow \sum\dfrac{y^3z}{x^2(xy+z^2)}\ge\dfrac{3}{2}$$
take 
$$a=\dfrac{x}{z},b=\dfrac{y}{x},c=\dfrac{z}{y}$$where $abc=1$
$$\Longleftrightarrow \sum \dfrac{b^2}{a+b}\ge \dfrac{3}{2}$$
use Cauchy-Schwarz inequality we have
$$\sum\dfrac{b^2}{a+b}\ge\dfrac{(a+b+c)^2}{\sum(a+b)}=\dfrac{a+b+c}{2}\ge\dfrac{3}{2}$$
