<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] Let $x=\frac{1}{a}$ and $y=\frac{1}{b}$. We want to show that $x+y=1$ and $x,y\geq 0$ imply
$$ y^2x\sqrt{4x^2+1}+x^2y\sqrt{4y^2+1}\leq \frac{\sqrt{2}}{4}. $$
Letting $x=\frac{1+t}{2},y=\frac{1-t}{2}$, this is equivalent to finding the maximum of 
$$ f(t) =(1-t^2)\left[ (1-t)\sqrt{1+(1+t)^2}+(1+t)\sqrt{1+(1-t)^2}\right]$$
(which is an even function) over $[-1,1]$. We have
$$ f(t) = (1-t^2)\sqrt{4-2t^2+2t^4+2(1-t^2)\sqrt{1+t^4}} $$
$$ f(\sqrt{t}) = \sqrt{2}(1-t)\sqrt{2-t+t^2+(1-t)\sqrt{1+t^2}}$$
and both $1-t$ and $2-t+t^2+(1-t)\sqrt{1+t^2}$ are positive and decreasing functions over $[0,1]$, so the maximum of $f(t)$ is attained at the origin. Indeed
$$2-\tan\theta+\tan^2\theta+(1-\tan\theta)\sqrt{1+\tan^2\theta} = \frac{2+\sqrt{2}\sin\left(\tfrac{\pi}{4}+\theta\right)}{2\sin^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)}$$
and the derivative of the RHS is 
$$ -\frac{1}{\sin^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}<0.$$
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