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 .

Let $\{x,y,z\}=\{a^2,b^2,c^2\},$ where $a\geq b\geq c>0$.
Thus, $a^2+b^2+c^2=3$ and by AM-GM, C-S, Rearrangement and AM-GM again we obtain:
$$\sum_{cyc}\frac{xy^2}{4y^3+3}=\sum_{cyc}\frac{xy^2}{2y^3+1+2y^3+2}\leq\sum_{cyc}\frac{xy^2}{3y^2+4\sqrt{y^3}}\leq$$
$$\leq\frac{1}{(3+4)^2}\sum_{cyc}xy^2\left(\frac{3^2}{3y^2}+\frac{4^2}{4\sqrt{y^3}}\right)=\frac{9}{49}+\frac{4}{49}\sum_{cyc}x\sqrt{y}=$$
$$=\frac{9}{49}+\frac{4}{49}\left(\sqrt{x}\sqrt{xy}+\sqrt{y}\sqrt{yz}+\sqrt{z}\sqrt{zx}\right)\leq\frac{9}{49}+\frac{4}{49}\left(a\cdot ab+b\cdot ac+c\cdot bc\right)=$$
$$=\frac{9}{49}+\frac{4b}{49}\left(a^2+ac+c^2\right)=\frac{9}{49}+\frac{4abc}{49}+\frac{4b(3-b^2)}{49}\leq\frac{9}{49}+\frac{4\cdot1}{49}+\frac{4\cdot2}{49}=\frac{3}{7}$$
and we are done!
