# Proving that the following inequality holds in positive numbers

Assume $$S=x_1+x_2+\cdots+x_n$$ where $x_i>0$. Prove the following inequality:$$\frac{S}{S-x_1}+\frac{S}{S-x_2}+\cdot+\frac{S}{S-x_n}\ge\frac{n^2}{n-1}$$with equality iff $x_1=x_2=\cdots=x_n$.

My attempt:

Using AM-GM we have$$\frac{S}{S-x_1}+\frac{S}{S-x_2}+\cdots+\frac{S}{S-x_n}\ge n\left(\frac{S}{S-x_1}\frac{S}{S-x_2}\cdots\frac{S}{S-x_n}\right)^{\dfrac{1}{n}}$$so it suffices to show that $$\left( \frac{S}{S-x_1}\frac{S}{S-x_2}\cdots\frac{S}{S-x_n} \right)^{\dfrac{1}{n}}\ge \frac{n}{n-1}\tag{*}$$

Question 1: How do I go any further or any other idea to prove this inequality?

Question 2: Does $(*)$ hold at all?

any ideas?

Hint: by the AM $\ge$ harmonic mean inequality:

$$\frac{\sum (S-x_k)}{n} \ge \frac{n}{\sum \dfrac{1}{S-x_k}}$$

What's left to note is that the LHS is $\displaystyle \frac{nS - S}{n}= \frac{n-1}{n} S\,$.

i will give you a hint: for $n=2$ we have to prove that $$\frac{S}{S-x_1}+\frac{S}{S-x_2}\geq 4$$ by Cauchy Schwarz we have $$\frac{S^2}{S(S-x_1)}+\frac{S^2}{S(S-x_2)}\geq \frac{(S+S)^2}{S(S-x_1+S-x_2)}=\frac{4S^2}{S^2}=4$$ through this way you Can prove your inequality!

Note that

$$\dfrac{S}{S-x_1}+\dfrac{S}{S-x_2}+...+\dfrac{S}{S-x_n}=\sum{\dfrac{S}{S-x_i}}=\sum\dfrac{1}{1-\frac{x_i}{S}}$$

and by HM-AM

$$\frac{n}{\sum{\dfrac{1}{1-\frac{x_i}{S}}}} \le \frac{\sum {1-\frac{x_i}{S}}}{n}=\frac{n-1}{n}$$

By C-S $$\sum_{k=1}^n\frac{S}{S-x_k}=S\sum_{k=1}^n\frac{1^2}{S-x_k}\geq\frac{Sn^2}{\sum\limits_{k=1}^n(S-x_k)}=\frac{Sn^2}{Sn-S}=\frac{n^2}{n-1}$$ Also, we can use Jensen.

Indeed, let $f(x)=\frac{1}{S-x}.$

Thus, $$f''(x)=\frac{2}{(S-x)^3}>0.$$ Hence, $$S\sum_{k=1}^n\frac{1}{S-x_k}\geq nS\cdot\frac{1}{S-\frac{S}{n}}=\frac{n^2}{n-1}.$$