<s>[INST] Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$

Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagrange multiplier method has not gotten me further either.
 [/INST] To get rid of some square roots and simplify, substitute $(a,b)\to (2x, 2y)$. The inequality can be written as:
Let $x,y >0$ and $\frac{1}{x}+\frac{1}{y} = 2$. Prove that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq (x+y)^2$$
In this case, from AM-GM we can see that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq \frac{x^2+1+2}{2}+\frac{y^2+1+2}{2} = \frac{x^2+y^2}{2}+3$$
It will be enough to prove:
$$\frac{x^2+y^2}{2}+3 \leq (x+y)^2$$
or 
$$x^2+y^2+4xy \geq 6$$
From AM-GM on the initial condition, we can see that $xy \geq 1$. Therefore
$$x^2+y^2+4xy \geq 6xy \geq 6$$
Equality occurs when $(x,y) = (1,1)$, so when $(a,b)=(2,2)$.
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