How can I solve this inequality using convexity? Given $a, b, c\ge  0$ and $x, y, z> 0$ and $a + b + c = x + y + z$. 
Show that $$a ^ 3 / x ^ 2 + b ^ 3 / y ^ 2 + c ^ 3 / z ^ 2 \ge a + b + c$$ prove inequality using convexity

By Holder
$$\left(\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\right)(x+y+z)^2\geq(a+b+c)^3$$ and we are done!
Also, since $f(x)=x^3$ is a convex function on $[0,+\infty),$ by Jensen we obtain:
$$\sum_{cyc}\frac{a^3}{x^2}=(x+y+z)\sum_{cyc}\frac{x}{x+y+z}\left(\frac{a}{x}\right)^3\geq$$
$$\geq(x+y+z)\left(\sum_{cyc}\frac{x}{x+y+z}\frac{a}{x}\right)^3=a+b+c.$$
