By AM-GM, we have $x^2 + y^2 \ge 2\Big(\frac{x+y}{2}\Big)^2=\frac{1}{2}(x+y)^2$. Setting $x=a+\frac{1}{a}$ and $y=b+\frac{1}{b}$ yields: \begin{align} &\Big(a+\frac{1}{a}\Big)^2+\Big(b+\frac{1}{b}\Big)^2\\ &\ge \frac{1}{2}\Big[\Big(a+\frac{1}{a}\Big)+\Big(b+\frac{1}{b}\Big)\Big]^2\\ &= \Big(a+b+\frac{1}{a}+\frac{1}{b}\Big)^2\\&=\frac{1}{2}\Big(1+\frac{1}{ab}\Big)^2\\ &\ge\frac{1}{2}\times (1+4)^2&\scriptsize{(see\ below)}\\ &=\frac{25}{2} \end{align} The reason that $\frac{1}{ab}\ge 4$ is because $$1=a+b\ge 2\sqrt{ab}\implies\sqrt{ab}\le\frac{1}{2}$$ $$\therefore\quad\frac{1}{ab}=\frac{1}{(\sqrt{ab})^2}\ge\frac{1}{\Big(\frac{1}{2}\Big)^2}=4$$