One of the inequalities

Relevant wiki: Titu's Lemma

$\begin{aligned} \left(a+\frac 1a\right)^2 + \left(b+\frac 1b\right)^2 & \ge \frac {\left(a+\frac 1a + b+\frac 1b\right)^2}{1+1} & \small \color{#3D99F6} \text{Using Titu's lemma} \\ & = \frac 12 \left(1+{\color{#3D99F6} \frac 1a + \frac 1b} \right)^2 & \small \color{#3D99F6} \text{Using Titu's lemma again} \\ & \ge \frac 12 \left(1+{\color{#3D99F6} \frac {(1+1)^2}{a+b}}\right)^2 \\ & \ge \frac 12 \left(1+ \frac 41\right)^2 \\ & = \boxed{12.5} \end{aligned}$

Equality occurs when $a=b=\frac 12$ .

Here's my solution:

Let $f(x)=(a+\frac{1}{a})^2$ . We know that $f(x)$ is a convex function. So, by Jensen's Inequality, we have $f(\frac{a+b}{2}) \leq \frac{f(a)+f(b)}{2}$ and hence $f(\frac{1}{2}) \times 2 \leq (a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ . So, the minimum value of the expression is $\frac{25}{4} \times 2=\boxed{12.5}$

Using A.M/ G.M.