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.

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 $(a,b)=(2,2)$.
