Prove $\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$ and $\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$. Let $a_1,a_2,\ldots ,a_n$, be positive real numbers and  S = $a_1+a_2+a_3+\cdots+a_n$. Use the Cauchy-Schwarz inequality to prove that 
$$\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$$ and $$\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$$
Hint: Apply the inequality to:


*

*$\sum \frac{S}{a_i}$ and $\sum \frac{a_i}{S}$ for one case,    

*and $\sum a_i(S-a_i)$ and $\sum \frac{a_i}{S-a_i}$ for the other case.


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 $(2)$
$$
\begin{align*}
\displaystyle\sum_{i=1}^n \frac{a_i}{S-a_i} &\geq \left(\frac{n^2}{n-1}\right) - n\\
&\geq \frac{n}{n-1}
\end{align*}
$$
