Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ 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
 A: 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 $$\frac{xy}{x^2+y^2}\le\frac12$$
A: 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*}
A: $$\sum_{cyc}\frac{x^3}{x^2+y^2}-\frac{x+y+z}{2}=\sum_{cyc}\left(\frac{x^3}{x^2+y^2}-\frac{x}{2}\right)=\sum_{cyc}\frac{x^3-xy^2}{2(x^2+y^2)}=$$
$$=\sum_{cyc}\left(\frac{x^3-xy^2}{2(x^2+y^2)}-\frac{x-y}{2}\right)=\sum_{cyc}\frac{y(x-y)^2}{2(x^2+y^2)}\geq0.$$
