Beware

Appying AM-GM inequality, we have $\sqrt{pq}\le\dfrac{p+q}{2}=\dfrac{1}{2} \Rightarrow pq\le\dfrac{1}{4}$

We have:

$\quad\left(p+\dfrac1p\right)^2+\left(q+\dfrac1q\right)^2$

$=p^2+q^2+\dfrac{1}{p^2}+\dfrac{1}{q^2}+4$

$\geq 2\sqrt{p^2q^2}+2\sqrt{\dfrac{1}{p^2}.\dfrac{1}{q^2}}+4$

$=2pq+\dfrac{2}{pq}+4$

$=2pq+\dfrac{1}{8pq}+\dfrac{15}{8pq}+4$

$\ge2\sqrt{2pq.\dfrac{1}{8pq}}+\dfrac{15}{8.\dfrac{1}{4}}+4=\dfrac{25}{2}$

The equality holds if and only if $p=q=\dfrac{1}{2}$ .

I guessed p=q=0.5. Result 12.5. Then substituted p=0.5+k and q=0.5-k and checked when is the minimum of the expression. It turned out to be k=-1.3, 0 and +1.3, but k is less than 0.5, so the only solution was k=0, p=q=0.5.