Prove that if $x_1,x_2,\ldots,x_n>0$, then $$(1+x_1)(1+x_1+x_2)\ldots(1+x_1+x_2+\ldots+x_n) \ge \sqrt{(n+1)^{n+1}}\sqrt{x_1x_2\ldots x_n}$$ I've tried using the AM-GM inequality: $$(1+x_1)(1+x_1+x_2)\ldots(1+x_1+x_2+\ldots+x_n) \ge (n+1)!\sqrt{x_1}\sqrt[3]{x_1x_2}\ldots\sqrt[n+1]{x_1x_2\ldots x_n}$$ But this obviously leads to nowhere.

I can try to prove this by using mathematical induction. If the inequality is true for $n=k$, then, if we prove that the inequality is true for $n=k+1$, everything's proved if we prove one case, e.g. when n=1.

When n=1, then it is obvious that the inequality holds: $$1+x \ge 2\sqrt{x}$$ So if n=k+1, then $$(1+x_1)(1+x_1+x_2)\ldots(1+x_1+x_2+\ldots+x_{k+1}) \ge \sqrt{(k+2)^{k+2}}\sqrt{x_1x_2\ldots x_{k+1}}$$

But we know that $$(1+x_1)(1+x_1+x_2)\ldots(1+x_1+x_2+\ldots+x_k) \ge \sqrt{(k+1)^{k+1}}\sqrt{x_1x_2\ldots x_k}$$

So we can try proving that $$(1+x_1+x_2+\ldots+x_{k+1})\sqrt{(k+1)^{k+1}}\sqrt{x_1x_2\ldots x_k} \ge \sqrt{(k+2)^{k+2}}\sqrt{x_1x_2\ldots x_{k+1}}$$

We can divide both sides by $\sqrt{x_1x_2\ldots x_k}$ (it's greater than $0$, so we can do this). We get: $$(1+x_1+x_2+\ldots+x_{k+1})\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^{k+2}}\sqrt{x_{k+1}}\tag{1}$$

We could use AM-GM here to try to continue: $$(k+2)\sqrt{x_1x_2\ldots x_{k+1}}\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^{k+2}}\sqrt{x_{k+1}}$$

Now we can divide everything by $\sqrt{x_{k+1}}$ (it's greater than $0$). We get: $$(k+2)\sqrt{x_1x_2\ldots x_k}\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^{k+2}}$$

Now we can divide everything by $k+2$ ($n \in \mathbb N$, so obviously $k \in \mathbb N \Rightarrow k>0$). We get: $$\sqrt{x_1x_2\ldots x_k}\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^k}$$

We could square both sides now (both sides are greater than $0$, so that changes nothing). We get: $$x_1x_2\ldots x_k(k+1)^{k+1} \ge (k+2)^k$$

But if k=1, then $$4x \ge 3$$ So I've reduced the LHS too much. Forget the steps after $(1)$. We can still prove that: $$(1+x_1+x_2+\ldots+x_{k+1})\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^{k+2}}\sqrt{x_{k+1}}$$

  • $\begingroup$ I think that mathematical induction could help here... $\endgroup$ – Salech Rubenstein Dec 22 '13 at 15:07

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 $$\prod_{i=1}^{n}(1+x_{1}+x_{2}+\cdots+x_{i})\ge\sqrt{(n+1)^{n+1}x_{1}x_{2}\cdots x_{n}}$$


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