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 .
 A: 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.
A: Using AM-GM, we have
$$4y^3+3=y^3+y^3+y^3+y^3+1+1+1\geq 7\left((y^3)^41^3\right)^{1/7}=7y^{12/7}.$$
So
$$\frac{xy^2}{4y^3+3}\leq\frac{xy^2}{7y^{12/7}}=\frac{xy^{2/7}}{7}.$$
Now, note by Holder's inequality that
$$\sum_{cyc}xy^{2/7}\leq (x+y+z)^{5/7}(xy+yz+zx)^{2/7}.$$
By Cauchy, $xy+yz+zx\leq \frac{(x+y+z)^2}{3}$.  Thus,
$$\sum_{cyc}xy^{2/7}\leq \frac{(x+y+z)^{9/7}}{3^{2/7}}.$$  So
$$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{\sum_{cyc}xy^{2/7}}{7}\leq \frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}.$$
Thus, $\sum_{cyc}\frac{xy^2}{4y^3+3}\le\frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}$ for all $x,y,z\ge 0$, where the equality case is $x=y=z=1$.  When $x+y+z=3$, we get
$$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{3^{9/7}}{7\cdot 3^{2/7}}=\frac{3}{7}.$$
The equality holds iff $x=y=z=1$.
More generally, for non-negative real numbers $x,y,z$, for parameters $a,b>0$, and for real exponents $m,n$ such that $$(m-1)(a+b)\le an\le m(a+b),$$ we have $$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{\sum_{cyc}xy^{\frac{m(a+b)-an}{a+b}}}{a+b}\le \frac{\left(\sum_{cyc}x\right)^{\frac{an-(m-1)(a+b)}{a+b}}\left(\sum_{cyc}xy\right)^{\frac{m(a+b)-an}{a+b}}}{a+b}\leq \frac{\left(\sum_{cyc}x\right)^{\frac{(m+1)(a+b)-an}{a+b}}}{3^{\frac{m(a+b)-an}{a+b}}(a+b)}.$$
The equality case is $x=y=z=1$.  In particular if, in addition, $x+y+z=3$, we get
$$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{3}{a+b}.$$
The equality holds iff $x=y=z=1$.
A: 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!
