Weird Inequality that seems to be true Is it true that:
$$\left (3x+\frac{4}{x+1}+\frac{16}{y^2+3}\right )\left (3y+\frac{4}{y+1}+\frac{16}{x^2+3}\right )\geq 81,\ \forall x,y\geq 0$$
I have proved that $3x+\frac{4}{x+1}+\frac{16}{x^2+3}= 9 +\frac{(x-1)^2 (3x^2+1)}{(x+1)(x^2+3)}, \ \forall x\geq 0$, but I did not succeed in proving the initial inequality.
Thanks, for the counterexample. This is for sure good:
$$\left (3x+\frac{4}{x+1}+\frac{8}{\sqrt{2(y^2+1)}}\right )\left (3y+\frac{4}{y+1}+\frac{8}{\sqrt{2(x^2+1)}}\right )\geq 81,\ \forall x,y\geq 0$$

This inequality is false, e.g. $x=2.5$, $y=0.5$. You will obtain a value of $79.9901<81$.
