Prove this equality $\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}$ let $x^3+y^3+z^3=3,x,y,z>0$ show that
$$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$
I have show that
let $x,y,z$ be positive numbers,such that $x+y+z=3$,prove that
$$\dfrac{x}{1+y^3}+\dfrac{y}{1+z^3}+\dfrac{z}{1+x^3}\ge\dfrac{3}{2}$$
 pf: use $AM-GM$ we have 
$$\dfrac{x}{1+y^3}=x-\dfrac{xy^3}{1+y^3}\ge x-\dfrac{xy^3}{2y^{3/2}}=x-\dfrac{xy^{3/2}}{2}$$
and,similarly
$$\dfrac{y}{1+z^3}\ge y-\dfrac{yz^{3/2}}{2},\dfrac{z}{1+x^3}\ge z-\dfrac{zx^{3/2}}{2}$$
Thus,it suffices to show that
$$xy^{3/2}+yz^{3/2}+zx^{3/2}\le 3$$
and it is known that
$$(a^3b^2+b^3c^2+c^3a^2)^2\le\dfrac{1}{3}(a^2+b^2+c^2)^3$$
seting $x=a^2,y=b^2,z=c^2$,by done!
But for this  problem :
$$y^2+5=y^2+1+1+1+1+1\ge 6y^{1/3}$$
and similarly
$$z^2+5\ge 6z^{1/3}, x^2+5\ge 6x^{1/3}$$
it suffices prove that
$$xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$$
with $x^3+y^3+z^3=3$,I use maple find   this is ($xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3$,with $x^3+y^3+z^3=3$) not true!,But  after I use maple find this
$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2},x^3+y^3+z^3=3$ is true!
and my other idea
$$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$
$$\Longleftrightarrow 2\sum x^2y^3+10\sum x^2y+50\sum x-5\sum x^2y^2-25\sum y^2-x^2y^2z^2\le 95$$
so I think my methods can't prove this problem, can someone use other methods show it? Thank you everyone.

By Cauchy-Schwarz inequality
$$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \sqrt{x^2 + y^2 + z^2}\sqrt{\frac{1}{(x^2+5)^2} + \frac{1}{(y^2 + 5)^2} + \frac{1}{(z^2 + 5)^2}}.$$
By AM-GM the RHS is less than
$$\frac{\frac{x^2 + y^2 + z^2}{6} + \frac{6}{(x^2 +5)^2} + \frac{6}{(y^2+5)^2} + \frac{6}{(z^2+5)^2}}{2} = \sum \left(\frac{x^2}{12} + \frac{3}{(x^2+5)^2}\right).$$
By multiplying everything out and factoring, one can prove that for $x \in [0,\sqrt[3]{3}]$
$$\frac{x^2}{12} + \frac{3}{(x^2+5)^2} \le \frac{2x^3 + 7}{54}.$$
Indeed, the inequality is equivalent with
$$(x-1)^2(4x^5 - x^4 + 34x^3 - 7x^2 + 52x + 26) \ge 0,$$
which is clearly true for $x \ge 0$.
After we have this result, we get that
$$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \frac{2(x^3+y^3+z^3) + 3\cdot 7}{54} = \frac{1}{2}.$$
