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.

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:
$$\left(\frac{\sqrt 2}{4}(a+b)^2\right)^2 \stackrel{(1)}{=} \frac 18 (a+b)^2(ab)^2 \stackrel{(2)}{\geq}2(a^2+b^2+2ab)$$
$$ \stackrel{(2)}{\geq}2(a^2+b^2+8)$$
