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.


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