# $\sum_{i=1}^n \frac{1}{a+S-x_i} \leq M$

Given integer $$n \geq 2$$ and positive real number $$a$$, find the smallest real number $$M = M (n, a)$$,

such that for any positive real numbers $$x_1, x_2, \ldots, x_n$$ with $$x_1x_2 \ldots x_n=1$$,

the following inequality hold: $$\displaystyle\sum_{i=1}^n \frac{1}{a+S-x_i} \leq M$$ where $$S=\displaystyle\sum_{i=1}^nx_i$$

My attempt :

By trying small values, we claim that the answer is Max$$\{\frac{1}{a}, \frac{n}{n-1+a}\}$$

If $$a<1$$, we choose $$x_1= x_2= \ldots= x_{n-1}=k, x_n=\frac{1}{k^{n-1}}$$

so $$\displaystyle\sum_{i=1}^n \frac{1}{a+S-x_i}$$

$$= (n-1)\left(\frac{1}{a+S-k}\right)+\left(\frac{1}{a+S-k^{n-1}}\right)$$

$$= \left(\frac{1}{a+(n-1)k}\right)+\left(\frac{n-1}{a+(n-2)k+\frac{1}{k^{n-1}}}\right)$$

when $$k$$ goes to $$\infty$$, $$\displaystyle\sum_{i=1}^n \frac{1}{a+S-x_i}$$ tends toward $$\frac{1}{a}$$

If $$a \geq 1$$, we choose $$x_1= x_2= \ldots= x_{n-1}=1$$

so $$\displaystyle\sum_{i=1}^n \frac{1}{a+S-x_i} = \frac{n}{n-1+a}$$

Please suggest how to proceed. Thank you.