Lemma 1 Let ${{x}_{1}},{{x}_{2}},\ldots ,{{x}_{n}},{{a}_{0}},{{a}_{1}},...,{{a}_{n}}$ be positive real numbers such that ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}=1,{{a}_{0}}={{a}_{n}}$. If $k\in \mathbb{N},{{S}_{r}}=x_{1}^{r}+x_{2}^{r}+...+x_{n}^{r}$,prove that ${{a}_{0}}{{S}_{k}}+{{a}_{1}}{{S}_{1+k}}+{{a}_{2}}{{S}_{2+k}}+...+{{a}_{n}}{{S}_{n+k}}\le \left( {{a}_{0}}+{{a}_{1}}+...+{{a}_{n}} \right)\frac{{{S}_{n+k}}+{{S}_{n+k-1}}}{2}$ Lemma 2 Let ${{x}_{1}},{{x}_{2}},\ldots ,{{x}_{n}}$ be positive real numbers such that ${{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}=1$. Prove that \[\sum\limits_{i=1}^{n}{x_{i}^{n}}(1+{{x}_{i}})\ge \frac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}+1 \right)}^{n}}{{x}_{i}}}}{{{2}^{n-1}}}\] From Lemma 1, Lemma 2 and inequality MA-MG result: \[\sum\limits_{i=1}^{n}{x_{i}^{n}}(1+{{x}_{i}})\ge \frac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}+1 \right)}^{n}}{{x}_{i}}}}{{{2}^{n-1}}}\ge \frac{n}{{{2}^{n-1}}}\prod\limits_{i=1}^{n}{(1+{{x}_{i}})}.\]

This problem is killed via homogenising both sides and using Muirhead's Inequality. Alternative proof: Assume without loss of generality that $x_1 \ge x_2 \ge ... \ge x_n$. So $x_1^n \ge x_2^n \ge ... \ge x_n^n$ and $1+x_1 \ge 1+x_2 \ge ... \ge 1+x_n$. By Chebyshev's Inequality, $LHS \ge \frac{(x_1^n+x_2^n+...+x_n^n)(n+x_1+x_2+...+x_n)}{n} \ge (\frac{x_1+x_2+...+x_n}{n})^n (n+x_1+x_2+...+x_n)$. (Power Mean Inequality) By AM-GM, $RHS \le \frac{n}{2^{n-1}}(\frac{n+x_1+x_2+...+x_n}{n})^n$. Thus $LHS \ge RHS \Leftrightarrow (\frac{x_1+x_2+...+x_n}{n})^n \ge (\frac{n+x_1+x_2+....+x_n}{2n})^{n-1}$ which is true as by AM-GM, $x_1+x_2+...+x_n \ge n$.

I want to point a solution using Holder and Cebisev : Let $s = x_1 + .. + x_n$. It is clear that $s \ge n$. From Holder we have that : $(\sum_{i=1}^n x_i^n(x_i + 1))(\sum_{i=1}^n x_i + 1)^{n - 1} \ge (\sum_{i=1}^n x_i(x_i + 1))^n$ Let $A$ be the left side of our initial inequality , then : $A \ge \frac{(\sum_{i=1}^n x_i(x_i + 1))^n}{(\sum_{i=1}^n x_i + 1)^{n - 1}}$ Now $(\sum_{i=1}^n x_i(x_i + 1))^n \ge (\frac{1}{n} (\sum_{i=1}^n x_i) (\sum_{i=1}^n (x_i + 1)))^n \ge s^n \prod_{i=1}^n (x_i + 1)$ (The last step is from Cebisev and AM -GM) So now we have to prove that : $\frac{s^n \prod_{i=1}^n (x_i + 1)}{(\sum_{i=1}^n x_i + 1)^{n - 1}} \ge \frac{n}{2^{n-1}} \prod_{i=1}^n (x_i + 1)$ Wich is equivalent with : $\frac{s^n}{(s + n)^{n -1}} \ge \frac{n}{2^{n-1}}$ wich is equivalent to : $s^{n} 2^{n -1} \ge (s + n)^{n -1} n$ wich is obvious because $s \ge n$ . So we are done .

My solution by AM-GM we have $\sum_{i=1}^n \frac{x_i^n (1 + x_i)}{\prod_{i=1}^n (1 + x_i)}$+$(n-1)\sum_{i=1}^n\frac{1+x_i}{2^n}$ $\geq $$\sum_{i=1}^n\frac{nx_i}{2^{n-1}}$ and then we can get $\sum_{i=1}^n \frac{x_i^n (1 + x_i)}{\prod_{i=1}^n (1 + x_i)}$ $\geq $$\sum_{i=1}^n\frac{(n+1)x_i}{2^n}$-$\frac{n(n-1)}{2^n}$ and use AM-GM again$\sum_{i=1}^n\frac{(n+1)x_i}{2^n}$-$\frac{n(n-1)}{2^n}$ $\geq $$\frac{n}{2^{n-1}}$ we are done