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This question already has an answer here:

Let $a_1,a_2,\dots,a_n$ $\in$ $\mathbb{R}^+$ and $a_1\cdot a_2\cdots a_n=1$, , prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$ I have tried factorising but it just lead me to extremily complicated equation that were extremily difficult to understand... Could someone help me?

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marked as duplicate by Arnaud D., Xander Henderson, Mike Earnest, Yanior Weg, Joshua Mundinger Apr 19 at 18:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $1+a\ge 2\sqrt{a}$ for $a\ge 0$. $\endgroup$ – Fan Sep 2 '16 at 8:13
  • $\begingroup$ And....? I don't see why that would help... $\endgroup$ – user361491 Sep 2 '16 at 8:14
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    $\begingroup$ Thus $LHS\ge 2^n\sqrt{a_1\cdot a_2 \cdots a_n}$ $\endgroup$ – Fan Sep 2 '16 at 8:15
  • $\begingroup$ Could you please detail how you got there? $\endgroup$ – user361491 Sep 2 '16 at 8:16
  • $\begingroup$ You replace $a$ by $a_1, a_2, \cdots, a_n$ and time them up $\endgroup$ – Fan Sep 2 '16 at 8:17
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applying the AM-GM inequality n-times we get $$(1+a_1)(1+a_2)\cdot...\cdot(1+a_n)\geq 2^n\sqrt{a_1a_2\cdot...\cdot a_n}=2^n$$ since $$a_1a_2\cdot...\cdot a_n=1$$

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We will use basic AM-GM inequality to solve this problem.

Proof of the basic form

$$(\sqrt{a}-\sqrt{b})^2 \ge 0 \implies a + b -2\sqrt{ab} \ge 0 \implies {a+b\over2} \ge \sqrt{ab}$$

General proof here


Your question

$${1+a_1\over 2} \ge \sqrt{a_1} \implies \color{red}{1 + a_1} \ge \color{blue}{2\sqrt{a_1}}$$ $${1+a_2\over 2} \ge \sqrt{a_2} \implies \color{red}{1 + a_2} \ge \color{blue}{2\sqrt{a_2}}$$ $${1+a_3\over 2} \ge \sqrt{a_3} \implies \color{red}{1 + a_3} \ge \color{blue}{2\sqrt{a_3}}$$

$$\vdots$$

$${1+a_n\over 2} \ge \sqrt{a_n} \implies \color{red}{1 + a_n} \ge \color{blue}{2\sqrt{a_n}}$$

Multiplying all the red things together, $\color{red}{(1+a_1)(1+a_2)(1+a_3) \cdots (1+a_n)}$ .

And the blue things, $\color{blue}{(2\sqrt{a_1})(2\sqrt{a_2})(2\sqrt{a_3}) \cdots (2\sqrt{a_n})}= \color{blue}{2^n\sqrt{a_1a_2a_3\cdots a_n}} \leftarrow \text{(why ?)}$

Combing it all together we get,

$$\color{green}{(1+a_1)(1+a_2)\cdot...\cdot(1+a_n)\geq 2^n\sqrt{a_1a_2\cdot...\cdot a_n}}$$

$$\color{green}{(1+a_1)(1+a_2)\cdot...\cdot(1+a_n)\geq 2^n} \leftarrow \text{(why ?)}$$ $$\color{red}{\star}\color{green}{\star}\color{blue}{\star}\color{yellow}{\star}\color{indigo}{\star}\color{red}{\star}\color{green}{\star}\color{blue}{\star}\color{yellow}{\star}\color{indigo}{\star}\color{red}{\star}\color{green}{\star}\color{blue}{\star}\color{yellow}{\star}\color{indigo}{\star}\color{red}{\star}\color{green}{\star}\color{blue}{\star}\color{yellow}{\star}\color{indigo}{\star}$$ And we are done. If still something isn't clear please ask me.


Not very beginner friendly but certainly a good read on inequalities

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    $\begingroup$ Nice answer, +1. I'd recommend you to read the meta questions about using many colors in answers, I believe most people here won't like them. $\endgroup$ – YoTengoUnLCD Sep 2 '16 at 20:36
  • $\begingroup$ @YoTengoUnLCD Why ? Sorry i did not know that. Thanks for telling me. Should i change everything to black ? $\endgroup$ – A---B Sep 2 '16 at 20:40
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    $\begingroup$ Well, it's mostly a problem for color-blind people, I believe (though not sure, I don't remember exactly what I read in those questions), that green-blue combinations were hard for people with that disability to differentiate. I wouldn't change it, just have that in mind. $\endgroup$ – YoTengoUnLCD Sep 2 '16 at 20:43
  • $\begingroup$ @YoTengoUnLCD thanks, i will keep that in mind. $\endgroup$ – A---B Sep 2 '16 at 20:44
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See HUYGEN’S INEQUALITY. It's a more general result.

enter image description here

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  • $\begingroup$ Any reasons for down voting? $\endgroup$ – rtybase Sep 2 '16 at 22:02
  • $\begingroup$ I can come up with reasons for downvotes but I find it a helpful pointer (+1 from me). $\endgroup$ – user66081 Sep 2 '16 at 22:09
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It's also Holder inequality for $n$ sequences.

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