Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$

Question

Help to prove this Inequality:

If x,y,z are postive real numbers then:

$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$

I tied to use analytic method with convex function but no result:

Since $f(x)=\frac{1}{x}$ is a convex function, by Jensen we obtain: $$\frac{1}{x+y+z}\sum_{cyc}\frac{x^3}{x^2+y^2}=\sum_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{1}{\frac{x^2+y^2}{x^2}}\right)\geq$$ $$\geq\frac{1}{\sum\limits_{cyc}\left(\frac{x}{x+y+z}\cdot\frac{x^2+y^2}{x^2}\right)}=\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}.$$ Thus, it's enough to prove that $$\frac{x+y+z}{\sum\limits_{cyc}\left(x+\frac{y^2}{x}\right)}\geq\frac{1}{2}$$ or $$x+y+z\geq\sum_{cyc}\frac{y^2}{x},$$ which is wrong.

thanks

Answer 1

$$\sum_{cyc}\frac{x^3}{x^2+y^2}-\frac{x+y+z}{2}=\sum_{cyc}\left(\frac{x^3}{x^2+y^2}-\frac{x}{2}\right)=\sum_{cyc}\frac{x^3-xy^2}{2(x^2+y^2)}=$$ $$=\sum_{cyc}\left(\frac{x^3-xy^2}{2(x^2+y^2)}-\frac{x-y}{2}\right)=\sum_{cyc}\frac{y(x-y)^2}{2(x^2+y^2)}\geq0.$$

Answer 2

Hint: We have $$ \frac {x^3}{x^2 + y^2}=x-\frac{xy^2}{x^2+y^2}\ge x-\frac{y}{2}$$ because by AM-GM $x^2+y^2\geq 2xy$ so that $$\frac{xy}{x^2+y^2}\le\frac12$$

Answer 3

We apply AM-GM Inequality with $x^2 + y^2 \geq 2xy$. \begin{align*} &\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \\ &= \dfrac{x^3 + xy^2}{x^2+y^2}+\dfrac{y^3 + yz^2}{y^2+z^2}+\dfrac{z^3 + zx^2}{z^2+x^2} - \left(\dfrac{xy^2}{x^2+y^2}+\dfrac{yz^2}{y^2+z^2}+\dfrac{zx^2}{z^2+x^2}\right) \\ &= x + y + z - \left(\dfrac{xy^2}{x^2+y^2}+\dfrac{yz^2}{y^2+z^2}+\dfrac{zx^2}{z^2+x^2}\right) \\ &\geq x + y + z - \left(\dfrac{xy^2}{2xy}+\dfrac{yz^2}{2yz}+\dfrac{zx^2}{2xz}\right) \\ &= x + y + z - \left(\frac{x}{2} + \frac{y}{2} + \frac{z}{2}\right) \\ &= \frac{x + y + z}{2} \end{align*}