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.
 A: 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 
$$ -\frac{1}{\sin^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\left(1+\tan\frac{\theta}{2}\right)}<0.$$
A: By AM-GM and C-S we obtain:
$$\frac{\sqrt2(a+b)^2}{4}=\frac{\sqrt{2(a+b)}\sqrt{a+b}(a+b)^2}{4ab}\geq(\sqrt{a}+\sqrt{b})\sqrt{a+b}=$$
$$=\sqrt{a^2+ab}+\sqrt{b^2+ab}\geq\sqrt{a^2+\frac{4a^2b^2}{(a+b)^2}}+\sqrt{b^2+\frac{4a^2b^2}{(a+b)^2}}=$$
$$=\sqrt{a^2+4}+\sqrt{b^2+4}.$$
A: 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 $(a,b)=(2,2)$.
A: You can show it just by squaring, using GM-AM and applying the two facts


*

*$(1)$: $a+b = ab$ and

*$(2)$: $ab \geq 4$
LHS:
$$\left(\sqrt{a^2+4}+\sqrt{a^2+4}\right)^2 =a^2+b^2+8 + 2\sqrt{(a^2+4)(b^2+4)}\stackrel{GM-AM}{\leq}2(a^2+b^2+8)$$
RHS:
$$\left(\frac{\sqrt 2}{4}(a+b)^2\right)^2 \stackrel{(1)}{=} \frac 18 (a+b)^2(ab)^2 \stackrel{(2)}{\geq}2(a^2+b^2+2ab)$$
$$ \stackrel{(2)}{\geq}2(a^2+b^2+8)$$
A: Hint:
You can use the constraint $\frac{1}{a}+\frac{1}{b}=1$ to


*

*solve $a$ in terms of $b$ and substitute 

*define a function in one variable $ f(b) \geq 0$
$$\cases{f(b) = \frac{\sqrt 2}{4}(a(b)+b)^2-\sqrt{a(b)^2+4}-\sqrt{b^2+4}\\\text{   where}\\ a(b) = \left(1-\frac{1}{b}\right)^{-1} = \frac{b}{b-1}}$$

*do one variable calculus on $f$ w.r.t. the remaining variable $b$.



If we do this, we can verify $f(2)=f'(2)=0$. The reason for why to calculate at 2 is that we can realize that $2$ is special just with symmetry, that $a=b=2$ should be special, since both constraint and inequality are symmetric w.r.t. a,b. $f(b) = f(\frac{b}{b-1})$.
What remains is to prove sub-interval $]1,2]$ (or $[2,\infty[$) is monotonic.
Here we should have done Jack's substitutions as it makes the calculations easier.
