# with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$

Let $x,y,z\ge 0$, show that
$$\dfrac{y}{xy+2y+1}+\dfrac{z}{yz+2z+1}+\dfrac{x}{zx+2x+1}\le\dfrac{3}{4}$$

$$\sum_{cyc}\dfrac{y}{xy+y+1}\le 1$$
$$(1-xyz)^2\ge 0$$
#### Solutions Collecting From Web of "with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$"
Note that by CS inequality,$$\frac{4y^2}{xy^2+2y^2+y} =\frac{(y+y)^2}{(xy^2+y^2)+(y^2+y)}\le \frac{y^2}{xy^2+y^2}+\frac{y^2}{y^2+y}=\frac{1}{x+1}+\frac{y}{y+1}$$
$$4\sum_{cyc}\frac{y}{xy+2y+1}\le \sum_{cyc}\frac1{x+1}+\sum_{cyc}\frac{x}{x+1}=3$$
P.S. You will need to handle the case $xyz=0$ separately, shouldnt be too difficult.