How can I solve this inequality using convexity?

Question

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

Answer 1

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: $$\sum_{cyc}\frac{a^3}{x^2}=(x+y+z)\sum_{cyc}\frac{x}{x+y+z}\left(\frac{a}{x}\right)^3\geq$$ $$\geq(x+y+z)\left(\sum_{cyc}\frac{x}{x+y+z}\frac{a}{x}\right)^3=a+b+c.$$