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#### melese

##### Member

- Feb 24, 2012

- 27

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

- Thread starter melese
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- Thread starter
- #1

- Feb 24, 2012

- 27

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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- Feb 24, 2012

- 27

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 $(a_1+1)\cdot(a_2+1)\cdots(a_n+1)\geq(1+1)(1+1) \cdots(1+1)=2^n$.

- Jan 25, 2013

- 1,225

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

$a_i>0$

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

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

and the proof is done

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