Show that an inequality holds for all positive real numbers $a, b$ such that $ab \geq 1$ I found the following question on a past international competition:
Show that:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right) \geq 16$
for all positive real numbers $a, b$ such that $ab\geq 1$.
I solved it in the following way:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)$
$\displaystyle =ab+2a^2+\frac{2a}{b+1}+2b^2+4ab+\frac{4b}{b+1}+\frac{2b}{a+1}+\frac{4a}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5ab+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 9+\frac{2(a^2+2ab+a+2b)+2(b^2+b+2ab+2a)+4}{(a+1)(b+1)}$ (from AM-GM we have that $a^2+b^2\ge 2ab \ge 2$)
$\displaystyle \ge 9+4(a+1)(b+1)+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
$\displaystyle \ge 13+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
However we have that $a^2+b^2\ge (a+b)*\sqrt{a^2b^2} \ge a+b$ (this is true for the well known inequality that $x1^2+x2^2+...+xn^2\ge (x1+x2+...+xn)*\sqrt[n]{x1x2...xn}$), $a^2+b^2\ge 2ab\ge 2$. Hence:
$2b^2+2a^2+ab\ge a+b+3$ so $2b^2+2a^2+4ab+2a+2b\ge 3ab+3a+3b+3$
So, we have that $\displaystyle \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\ge 13+\frac{3ab+3a+3b+3}{(a+1)(b+1)}\ge 13+\frac{3(a+1)(b+1)}{(a+1)(b+1)} \ge 16$
I believe that my solution is correct, however I'm not completely certain, so could you please have a look at it and also share if there is an easier and simpler way of solving the problem?
 A: By AM-GM and C-S we obtain:$$\prod_{cyc}\left(a+2b+\frac{2}{a+1}\right)=\prod_{cyc}\left(\frac{a+1}{2}+\frac{2}{a+1}+2b+\frac{a}{2}-\frac{1}{2}\right)\geq$$
$$\geq\prod_{cyc}\left(2+2b+\frac{a}{2}-\frac{1}{2}\right)=\prod_{cyc}\left(2b+\frac{a}{2}+\frac{3}{2}\right)\geq$$
$$\geq\left(2\sqrt{ab}+\frac{1}{2}\sqrt{ab}+\frac{3}{2}\right)^2\geq\left(2+\frac{1}{2}+\frac{3}{2}\right)^2=16.$$
A: From the calculation, you have done, we have:
$f(a, b) = 2(a + b)^{2} + ab + \left[\frac{2a}{1 + b} + \frac{4b}{1 + b} + \frac{2b}{1 + a} + \frac{4a}{1 + a} + \frac{4}{(1 + a)(1 + b)}\right]$
$ = 2(a + b)^{2} + ab + \frac{\left [(1 + a) + (1 + b) \right]^{2}}{(1 + a)(1 + b)} + \frac{\left [a^{2} +b^{2} + 6ab + 2a + 2b \right]}{(1 + a)(1 + b)}
$
We know from the AM-GM inequality we have $2(a + b)^{2} \geq 8$ and $\frac{\left [(1 + a) + (1 + b) \right]^{2}}{(1 + a)(1 + b)} \geq 4$ and by the hypothesis $ab \geq 1$.
It follows that $f(a, b) \geq 8 + 1 + 4 + \frac{\left [a^{2} +b^{2} + 6ab + 2a + 2b \right]}{(1 + a)(1 + b)}$ and if we can bound from above the last term by $3$ we would be done.
We have: $a^{2} +b^{2} + 6ab + 2a + 2b - 3(1 + a)(1 + b) = a^{2} + b^{2} + 3ab - 3 - a - b$
$ \geq a(a - 1) + b(b -1)\geq b^{-1}(b^{-1} -1) + b(b-1)$
$= b^{-2}[(1-b) + b^{3}(b-1)] = b^{-2}(b-1)[b^{3} - 1] $
$= b^{-2}(b-1)^{2} (b^{2} + b + 1) > 0$.
The desired bound is reached, as it was promised.
