# 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:

$$\left (3x+\frac{4}{x+1}+\frac{8}{\sqrt{2(y^2+1)}}\right )\left (3y+\frac{4}{y+1}+\frac{8}{\sqrt{2(x^2+1)}}\right )\geq 81,\ \forall x,y\geq 0$$

• Would differentiating the original expression wrt $x$ and then $y$ to find the minimum value of that expression help? – Shuri2060 Mar 3 '16 at 21:52
• Afraid not: Try $x=2.5$ and $y=0.5$. – Pixel Mar 3 '16 at 21:55
• Have you written the question correctly? Should all the terms in the first bracketted expression involve only $x$ and those in the second one only $y$? – Mufasa Mar 3 '16 at 22:00

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