Cauchy Schwarz Inequality:
$ \begin{align*} \left( \frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_n}\right)\left( a_1+a_2+\dots+a_n \right) &\geq \left(\frac{1}{\sqrt{a_1}}\sqrt{a_1}+\frac{1}{\sqrt{a_2}}\sqrt{a_2}+\dots +\frac{1}{\sqrt{a_n}}\sqrt{a_n} \right)^2 \\\ &\geq \left( 1+1+\dots+1\right)^2 = n^2 \end{align*} $
which when applied here $ \begin{align*} \displaystyle\sum_{i=1}^n \frac{S}{a_i} &= \left( \frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_n}\right)S \geq n^2\\ \end{align*} $
Therefore $ \begin{align*} \displaystyle\sum_{i=1}^n \frac{S-a_i}{a_i} &= \displaystyle\sum_{i=1}^n \frac{S}{a_i} -n\\ &\geq n^2 - n = n(n-1) \tag{1} \end{align*} $
Similarly, re-writing the other summation
$ \begin{align*} \displaystyle\sum_{i=1}^n \frac{a_i}{S-a_i} &= \frac{S-\displaystyle\sum_{j \neq 1}a_j}{\displaystyle\sum_{j \neq 1}a_j} + \frac{S-\displaystyle\sum_{j \neq 2}a_j}{\displaystyle\sum_{j \neq 2}a_j}+\dots +\frac{S-\displaystyle\sum_{j \neq n}a_j}{\displaystyle\sum_{j \neq n}a_j}\\ &= \displaystyle\sum_{i=1}^n \left(\frac{S}{\displaystyle\sum_{j \neq i}a_j}\right)-n \tag{2} \end{align*} $
Also since
$ \displaystyle\sum_{j \neq 1}a_j+\displaystyle\sum_{j \neq 2}a_j+\dots+\displaystyle\sum_{j \neq n}a_j = (n-1)S $
By Cauchy-Schwarz Inequality $ \begin{align*} \displaystyle\sum_{i=1}^n \left(\frac{1}{\displaystyle\sum_{j \neq i}a_j}\right)(n-1)S \geq n^2\\ \Longrightarrow \displaystyle\sum_{i=1}^n \left(\frac{1}{\displaystyle\sum_{j \neq i}a_j}\right) \geq \frac{n^2}{(n-1)S} \end{align*} $
From $(2)$
$ \begin{align*} \displaystyle\sum_{i=1}^n \frac{a_i}{S-a_i} &\geq \left(\frac{n^2}{n-1}\right) - n\\ &\geq \frac{n}{n-1} \end{align*} $