Prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$ 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?
 A: 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 $$a_1a_2\cdot...\cdot a_n=1$$
A: 
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 
A: See HUYGEN’S INEQUALITY. It's a more general result. 

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