<s>[INST] 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 .
 [/INST] 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 $x=y=z=1$.
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