$AM-GM$-ish inequality

Question

Suppose $x_0, \cdots, x_n$ are positive reals. Suppose that: $$\sum_{i = 0}^n \frac{1}{1+x_i} \leq 1$$ Then show that: $$\prod_{i=0}^{n} x_i \geq n^{n+1} $$

I got to this problem by rewriting problem $665$ in Andreescu's book. This feels like something pretty standard, but I can't figure it out.

Answer 1

Let $y_i:=\dfrac{1}{1+x_i}$ for $i=0,1,2,\ldots,n$. Then, $y_0,y_1,y_2,\ldots,y_n$ are positive real numbers such that $\sum\limits_{i=0}^n\,y_i\leq 1$. We want to prove that $$\prod_{i=0}^n\,\left(\frac{1-y_i}{y_i}\right)\geq {n^{n+1}}.$$

Let $[n]:=\{0,1,2,\ldots,n\}$. To prove the last inequality, we note that $$1-y_i\geq \sum_{j\in[n]\setminus\{i\}}\,y_j\geq n\,\left(\prod_{j\in[n]\setminus\{i\}}\,y_j\right)^{\frac{1}{n}}$$ for each $i\in[n]$. Thus, $$\prod_{i=0}^n\,\left(\frac{1-y_i}{y_i}\right)\geq \prod_{i=0}^n\,\left(\frac{n\,\left(\prod\limits_{j\in[n]\setminus\{i\}}\,y_j\right)^{\frac{1}{n}}}{y_i}\right)={n^{n+1}}\,.$$ The equality holds if and only if $y_0=y_1=y_2=\ldots=y_n=\dfrac1{n+1}$.

Answer 2

Because by AM-GM $$\prod_{i=0}^n\frac{x_i}{1+x_i}\geq\prod_{i=0}^n\sum_{k\neq i}\frac{1}{1+x_k}\geq\prod_{i=0}^n\left(n\prod_{k\neq i}\frac{1}{\sqrt[n]{1+x_k}}\right)=\frac{n^{n+1}}{\prod\limits_{i=0}^n(1+x_i)}$$ and we are done!