Inequality Problem: $(a_1+1)(a_2+1)...(a_n+1)\geq2^n$

[melese]

Here's a nice problem.
Suppose $a_1,a_2,...,a_n$ are postive real numbers satisfying \(a_1\cdot a_2\cdots a_n=1\). Show that $(a_1+1)(a_2+1)\cdots(a_n+1)\geq2^n$.

 

[tarantella]

Let $I_n:=\{1,\ldots,n\}$. Then $$\prod_{j=1}^n(1+a_j)=\sum_{J\subset I_n}\prod_{j\in J}a_j=\sum_{j\in J}\left(\prod_{j\in J}a_j\right)^{-1}$$ so $2\prod_{j=1}^n(1+a_j)=\sum_{J\subset I_n}f\left(\prod_{j\in J}a_j\right)$, where $f(x)=x+\frac 1x$. Since $f(x)-2=\frac{x^2+1-2x}x=\frac{(x-1)^2}x\geq 0$, we have $$2\prod_{j=1}^n(1+a_j)\geq \sum_{J\subset I_n}2=2\cdot 2^n$$ so $\prod_{j=1}^n(1+a_j)\geq 2^n$. We notice that we have equality if and only if for all $J\subset I_n$ we have $\prod_{j\in J}a_j=1$ i.e. for all $j\in I_n$, $a_j=1$.

 

[melese]

Here's another solution.

In general, if we consider the postive number pairs $(t,s)$ and $(\sqrt {ts},\sqrt {ts})$, we find that \[ts=\sqrt {ts}\sqrt {ts}\], but by the AM-GM inequality \[(\sqrt {ts}+1)(\sqrt {ts}+1)=(\sqrt {ts}+1)^2=ts+2\sqrt {ts}+1\leq ts+2\frac{t+s}{2}+1=(t+1)(s+1)\] with equality only when $t=s$. This means that the product $(t+1)(s+1)$ is minimal exactly when $t=s$ to begin with.

If not all the $a_i$ are equal, then WLOG $a_1\neq a_2$. But then, \[(a_1+1)\cdot(a_2+1)\cdots(a_n+1)\geq(\sqrt{a_1a_2}+1)(\sqrt{a_1a_2}+1)\cdots(a_n+1),\] which means that minimum value of $(a_1+1)\cdot(a_2+1)\cdots(a_n+1)$ is obtained only if we assume all the $a_i$ are equal. The assumption $a_1\cdot a_2\cdots a_n=1$ gives $a_1=a_2=...=a_n=1$.

So it's true that $(a_1+1)\cdot(a_2+1)\cdots(a_n+1)\geq(1+1)(1+1) \cdots(1+1)=2^n$.

 

[awkward]

Another approach:

Consider the function f(x) = ln(1 + e^x). Show that the second derivative is positive, hence f is convex. Then apply Jensen's Inequality.

 

[Albert1]

since for i=1,2,3,-----n

$a_i>0$

$a_i+1\geq 2\sqrt {a_i\times1}$

$\therefore (a_1+1)(a_2+1)--------(a_n+1)\geq 2^n(\sqrt{a_1a_2a_3------a_n}=2^n$

and the proof is done