Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$

From the hypothesis: $a+b=2ab$
So the claim can be written as: $2ab+\frac{1}{1+\sqrt{ab}}\geqslant 2.5$
Now to make everything based on $ab$, we analyze the hypothesis:
For which real values $x$, there exist $a,b$ such that $ab=x , \frac{1}{a}+\frac{1}{b}=2$?
For solving this put $a$ as our variable. We want to find $a$ such that $\frac{1}{a}+\frac{a}{x}=2$ or equivalently: $2ax=x+a^2$ now if we see this as a polynomial, $a$ exists iff $4x^2-4x\geqslant 0$ which happens iff $x\geqslant 1$ or $0\geqslant x$.
So now we can state the problem as follows:
For any real number $x$ such that $x$ isn't between $0$ and $1$, prove that $2x+\frac{1}{1+\sqrt x }\geqslant 2.5$
Now note that $x$ can't be negative because of the square root and this function is increasing for $x\geqslant 1$ so the minimum value is taken on $x=1$ which completes the proof.
