Hard problem

First of all, note that $\displaystyle\sqrt{ \frac{a^2+b^2}{2} } \ge \frac{a+b}{2}$ $\displaystyle (ab)^2\ge \left(\frac{a+b}{2}\right)^4$

both actually work $\forall a,b\in\mathbb R$ . You can see this for yourself, I won't present a proof, although I have rigorous proofs for these.

---

$\displaystyle \sqrt{ \frac{a^2+b^2}{2} } \ge \frac{a+b}{2}= \frac{1}{2}$ $\displaystyle \implies a^2+b^2\ge 0.5\tag{1}$ $\displaystyle \sqrt{ \frac{\left( \frac{1}{a}\right)^2+\left( \frac{1}{b} \right)^2 }{2} }\ge \frac{\frac{1}{a}+\frac{1}{b}}{2}$

$\displaystyle \implies \left(\frac{1}{a}\right)^2+\left( \frac{1}{b} \right)^2\ge \frac{1}{2(ab)^2}$ $\displaystyle \ge \cfrac{1}{2\left(\cfrac{a+b}{2}\right)^4}=8\tag{2}$ $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$

$\displaystyle =a^2+b^2+\left(\frac{1}{a} \right)^2+\left(\frac{1}{b} \right)^2+4\stackrel{\text{(1)(2)}}\ge \boxed {12.5}$

Set $f(x)=(x+\frac{1}{x})^{2}$ , we have $f''(x)=2(1+\frac{3}{x^{4}})\geq 0$

So $f(x)$ is convex and $\frac{f(a)+f(b)}{2} \geq f(\frac{a+b}{2})=\frac{25}{4}$