Show that $\frac{2}{(x + y + z)} \left(\frac{x ^ 3}{x ^ 2 + y ^ 2} + \frac{y ^ 3}{y ^ 2 + z ^ 2} + \frac{z ^ 3}{z ^ 2 + x ^ 2}\right) \geq 1$

Question

Question

If $x, y, z > 0$ show that:

$$\frac{2}{(x + y + z)} \left(\frac{x ^ 3}{x ^ 2 + y ^ 2} + \frac{y ^ 3}{y ^ 2 + z ^ 2} + \frac{z ^ 3}{z ^ 2 + x ^ 2}\right) \geq 1$$

My idea

If we divide by $\frac{2}{(x + y + z)}$ we have to show that:

$$\left(\frac{x ^ 3}{x ^ 2 + y ^ 2} + \frac{y ^ 3}{y ^ 2 + z ^ 2} + \frac{z ^ 3}{z ^ 2 + x ^ 2}\right) \geq \frac{x+y+z}{2}$$

from here I tried using the inequality of means or CBS but I got to nothing useful.

I don't know where to start and I always had this problem when I had to demonstrate inequality. Hope one fo you can help me! Thank you!

Answer 1

We have $$\frac{x^3}{x^2 + y^2} - x = \frac{-xy^2}{x^2 + y^2} = (-y/2) \cdot \frac{2xy}{x^2 + y^2} \ge - y/2.$$ Thus, we have $$\frac{x ^ 3}{x ^ 2 + y ^ 2} + \frac{y ^ 3}{y ^ 2 + z ^ 2} + \frac{z ^ 3}{z ^ 2 + x ^ 2} \ge (x - y/2) + (y - z/2) + (z - x/2).$$ We are done.