Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :

Let $x,y,z>0$ such that $x+y+z=3$ then we have :
  $$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$

I try to use Jensen's inequality but the function $f(x)=\frac{x^2}{4x^3+3}$ is neither concave or convex on the interval $[0,3]$
I can't use Karamata's inequality too .
Maybe brute force is the only way to solve it . 
I try also to use the derivative but it becomes a little bit difficult . 
In fact my idea was to use rearrangment inequality we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{x^3}{4x^3+3}+\frac{y^3}{4y^3+3}+\frac{z^3}{4z^3+3}$$
An use the inequality of Jensen's on $[0.8,1.2]$ with $f(x)=\frac{x^3}{4x^3+3}$
So it's a partial answer .
My question is how to complete my answer or can you provide an other answer ?
Thanks a lot for sharing your knowledge and your time .

Note that (tangent line trick)
$$\frac{5+2y}{49} - \frac{y^2}{4y^3+3}  = \frac{(8y^2+36y+15)(y-1)^2}{49(4y^3+3)}.$$
Thus, we have
$$\frac{y^2}{4y^3+3} \le \frac{5+2y}{49}, \quad \forall y \ge 0.$$
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{xy^2}{4y^3+3} \le \sum_{\mathrm{cyc}} \frac{x(5+2y)}{49}
= \frac{5(x+y+z) + 2(xy+yz+zx)}{49} \le \frac{3}{7}$$
where we have used the fact that $xy+yz+zx \le \frac{(x+y+z)^2}{3}$. We are done.
