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 $\frac{3}{2}$
On
By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x^3yz}{z^3(y^2+xz)}=\sum_{cyc}\frac{x^4y^2}{z^2xy(y^2+xz)}\geq\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{\sum\limits_{cyc}z^2xy(y^2+xz)}=$$ $$=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{xyz\sum\limits_{cyc}(y^2z+z^2x)}=\frac{\left(\sum\limits_{cyc}x^2y\right)^2}{2xyz\sum\limits_{cyc}x^2y}=\frac{1}{2}\sum_{cyc}\frac{x}{z}\geq\frac{3}{2}.$$
$$\Longleftrightarrow \sum\dfrac{y^3z}{x^2(xy+z^2)}\ge\dfrac{3}{2}$$ 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 $$\sum\dfrac{b^2}{a+b}\ge\dfrac{(a+b+c)^2}{\sum(a+b)}=\dfrac{a+b+c}{2}\ge\dfrac{3}{2}$$