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 $\frac{1}{a}+\frac{1}{b}=2?$
 A: We have: $4(ab)^2 = (a+b)^2 \ge 4ab \implies ab \ge 1\implies LHS \ge 2\sqrt{ab} + \dfrac{1}{1+\sqrt{ab}}= f(t), t = \sqrt{ab} \ge 1\implies f'(t) = 2 - \dfrac{1}{(1+t)^2}> 0\implies f(t) \ge f(1) = 2+\dfrac{1}{2} = \dfrac{5}{2} \implies LHS \ge \dfrac{5}{2} = RHS $. Equality occurs when $a = b = 1$.
A: 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.
A: Here is a solution that does not involve derivatives.
By AM-GM, we have that $a+b \geq 2\sqrt{ab}$. Hence it suffices for us to prove that $2\sqrt{ab}+\dfrac{1}{1+\sqrt{ab}} \geq \dfrac{5}{2}.$
From the given condition, $\dfrac{1}{a} + \dfrac{1}{b} =2 \Rightarrow a+b=2ab \Rightarrow 2ab \geq 2\sqrt{ab} \Rightarrow \sqrt{ab} \geq 1$. Hence, letting $x=\sqrt{ab}, x \geq 1$, we have to prove that:
\begin{align}
& 2x+\dfrac{1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x(1+x)+1}{1+x} \geq \dfrac{5}{2} \\
& \iff \dfrac{2x^2+2x+1}{1+x} \geq \dfrac{5}{2} \\
& \iff 2(2x^2+2x+1) \geq 5+5x \\
& \iff 4x^2-x \geq  3 \\
& \iff x(4x-1) \geq 3 \\
\end{align}
But the last inequality is obvious since $x \geq 1$ and $4x-1 \geq 4-1=3$; hence we are done.
