G4.2 Given that $a$ and $b$ are positive real numbers and $a+b=2$. If $S=\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$, find the minimum value $S$.
$$
\begin{array}{l}
(\sqrt{a}-\sqrt{b})^{2} \geq 0 \Rightarrow a+b-2 \sqrt{a b} \geq 0 \Rightarrow 1 \geq \sqrt{a b} \Rightarrow 1 \geq a b \\
a^{2}+b^{2}=(a+b)^{2}-2 a b=4-2 a b \geq 4-2=2 \ldots \ldots \ldots \ldots . .(2) \\
\frac{1}{a b} \geq 1 \Rightarrow \frac{1}{a^{2} b^{2}} \geq 1 \ldots \ldots \ldots \text { (3) } \\
S=\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+4 \\
=a^{2}+b^{2}+\frac{a^{2}+b^{2}}{a^{2} b^{2}}+4 \\
=\left(a^{2}+b^{2}\right)\left(1+\frac{1}{a^{2} b^{2}}\right)+4 \\
\geq 2 \times(1+1)+4=8 \text { (by (2) and (3)) }
\end{array}
$$
$\qquad$