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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?


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

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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

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 $$a_1a_2\cdot...\cdot a_n=1$$


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

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


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

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

See HUYGEN’S INEQUALITY. It's a more general result.

enter image description here

  • $\begingroup$ 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

It's also Holder inequality for $n$ sequences.