<s>[INST] 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 $\frac{1}{a}+\frac{1}{b}=2?$
 [/INST] 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.
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