The answer is 12.5.

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Expanding the expression, we have:

$\begin{aligned} \left(a+\frac 1a\right)^2 + \left(b+\frac 1b\right)^2 & = a^2 + 2 + \frac 1{a^2} + b^2 + 2 + \frac 1{b^2} \\ \blue{a^2+b^2} + 4 + \blue{\frac 1{a^2}+\frac 1{b^2}} & \ge \blue{\frac 12 (a+b)^2} + 4 + \blue{\frac 12 \left(\frac 1a+ \frac 1b\right)^2} & \small \blue{\text{By Titu's lemma}} \\ & = \blue{\frac 12} + 4 + \frac 12 \left(\red{\frac 1a+ \frac 1b}\right)^2 & \small \red{\text{By AM-HM inequality}} \\ & \ge \frac 12 + 4 + \frac 12 \left(\red{\frac 4{a+b}}\right)^2 \\ & = \frac 12 + 4 + \frac 12 \left(\red 4\right)^2 \\ & = \boxed{12.5} \end{aligned}$

For both the Titu's lemma and AM-HM inequality cases above, equality occurs when $a=b=\frac 12$ .