a and b are real positive number

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 $a=b=\frac 12$ .

This problem can be done in more than one way. One method would be to use the method of Lagrange multipliers while the other involves converting a multivariable problem to a single-variable problem. Here, the single variable approach is presented. The expression to be minimised is:

$y =\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 \implies y = \left(a + \frac{1}{a}\right)^2 + \left(1-a + \frac{1}{1-a}\right)^2 \because a+b =1$

Differentiating $y$ with respect to $a$ , equating the derivative of $y$ to zero, and after simplifying, gives the expression:

$a - \frac{1}{a^3} = (1-a) - \frac{1}{(1-a)^3}$

Looking at the similarity of the expressions on the left and right side leads to a trivial conclusion (there might be non-trivial conclusions):

$a = 1-a \implies a = b \implies a=b=\frac{1}{2} \because a+b =1$

From here, the expression using the computed values evaluates to $\boxed{\red{y_{min} = 12.5}}$ . To be more rigorous, one must carry out the second derivative test to confirm a minimum. This has been left out in this brief solution. This analysis is quite incomplete as it does not consider the existence of multiple local optimum points.