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

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The discussion centers around proving the inequality \((a_1+1)(a_2+1)\cdots(a_n+1) \geq 2^n\) for positive real numbers \(a_1, a_2, \ldots, a_n\) satisfying \(a_1 \cdot a_2 \cdots a_n = 1\). The proof employs the AM-GM inequality and properties of convex functions, demonstrating that equality holds only when all \(a_i\) are equal to 1. Various approaches are explored, including using logarithmic functions and Jensen's Inequality to establish the convexity of the function involved. The conclusion reinforces that the minimum value occurs when all \(a_i\) are equal, confirming the inequality. Thus, the inequality is validated under the given conditions.
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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 $(a_1+1)(a_2+1)\cdots(a_n+1)\geq2^n$.
 
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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 $j\in I_n$, $a_j=1$.
 
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 $(a_1+1)\cdot(a_2+1)\cdots(a_n+1)\geq(1+1)(1+1) \cdots(1+1)=2^n$.
 
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.
 
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
 
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