Let $x_1, x_2, \dots, x_n$ be nonnegativ real numbers.

(a) Prove that if $x_1x_2\dots x_n=1$, then $$\sum_{i=1}^n \dfrac{1}{n-1+x_i}\leq 1$$

(b) Find the maximum value of $$\sum_{i=1}^n \dfrac{1}{\frac{n}{x_i}-1+x_i^n}$$ where $x_1+x_2+\dots +x_n=n$.

I rewrote the expression: $$\sum_{i=1}^n \dfrac{1}{1+y_i} \leq n-1$$ where $$y_i=\dfrac{x_i}{n-1}$$

  • 2
    $\begingroup$ What is your question, exactly? What did you get when you rewrote the expression, and how is that related to the question? $\endgroup$ – Steve Kass Jan 15 '18 at 20:59

a) By AM-GM $$\sum_{i=1}^n\frac{1}{n-1+x_i}=\frac{1}{n-1}\sum_{i=1}^n\left(1-\frac{x_i}{x_i+n-1}\right)=$$ $$=\frac{n}{n-1}-\frac{1}{n-1}\sum_{i=1}^n\frac{x_i}{x_i+(n-1)\sqrt[n]{\prod\limits_{k=1}^nx_k}}=$$ $$=\frac{n}{n-1}-\frac{1}{n-1}\sum_{i=1}^n\frac{\sqrt[n]{x_i^{n-1}}}{\sqrt[n]{x_i^{n-1}}+(n-1)\sqrt[n]{\prod\limits_{k\neq i}x_k}}\leq$$ $$\leq\frac{n}{n-1}-\frac{1}{n-1}\sum_{i=1}^n\frac{\sqrt[n]{x_i^{n-1}}}{\sqrt[n]{x_i^{n-1}}+\sum\limits_{k\neq i}\sqrt[n]{x_k^{n-1}}}=\frac{n}{n-1}-\frac{1}{n-1}\cdot1=1.$$

b) By AM-GM $$\sum_{i=1}^n\frac{1}{\frac{n}{x_i}+x_i^n-1}\leq\sum_{i=1}^n\frac{1}{(n+1)\sqrt[n+1]{\left(\frac{1}{x_i}\right)^n\cdot x_i^n}-1}=1.$$ The equality occurs when all $x_i=1$, which says that $1$ is a maximal value.


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