19
$\begingroup$

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: $$(1+x_1+x_2+\ldots+x_{k+1})\sqrt{(k+1)^{k+1}} \ge \sqrt{(k+2)^{k+2}}\sqrt{x_{k+1}}$$

$\endgroup$
  • $\begingroup$ I think that mathematical induction could help here... $\endgroup$ – Salech Rubenstein Dec 22 '13 at 15:07
18
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.