Prove that if $x_1,x_2,\ldots,x_n>0$, then $(1+x_1)(1+x_1+x_2)\ldots(1+x_1+x_2+\ldots+x_n) \ge \sqrt{(n+1)^{n+1}}\sqrt{x_1x_2\ldots x_n}$.

Solution 1:

Let $$y_{1}=\dfrac{x_{1}}{1+x_{1}},y_{2}=\dfrac{x_{2}}{(1+x_{1})(1+x_{1}+x_{2})},y_{3}=\dfrac{x_{3}}{(1+x_{1}+x_{2})(1+x_{1}+x_{2}+x_{3})}\cdots,$$ $$y_{n}=\dfrac{x_{n}}{(1+x_{1}+\cdots+x_{n-1})(1+x_{1}+\cdots+x_{n})}, y_{n+1}=\dfrac{1}{1+x_{1}+\cdots+x_{n-1}+x_{n}}$$ then we have $$y_{1}+y_{2}+y_{3}+\cdots+y_{n+1}=\dfrac{x_{1}}{1+x_{1}}+\left(\dfrac{1}{1+x_{1}}-\dfrac{1}{1+x_{1}+x_{2}}\right)+\cdots=1$$ so Use AM-GM inequality,we have $$y_{1}+y_{2}+\cdots+y_{n}+y_{n+1}\ge (n+1)\sqrt[n+1]{y_{1}y_{2}\cdots y_{n}y_{n+1}}$$ then $$1\ge (n+1)\sqrt[n+1]{y_{1}y_{2}\cdots y_{n}y_{n+1}}=(n+1)\sqrt[n+1]{\dfrac{x_{1}x_{2}\cdots x_{n}}{(\prod_{i=1}^{n}(1+x_{1}+x_{2}+\cdots+x_{i}))^2}}$$ then we have $$\prod_{i=1}^{n}(1+x_{1}+x_{2}+\cdots+x_{i})\ge\sqrt{(n+1)^{n+1}x_{1}x_{2}\cdots x_{n}}.$$