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How do I prove that if $x_1, \ldots, x_n$ are positive real numbers, then $$1 \leq x_1 x_2 \cdots x_n \text{ implies that } 2^{n} \leq (1 + x_1)(1+x_2) \cdots (1 + x_n).$$

I attempted a proof by induction but am not able to nail the inductive step. Any help would be appreciated!

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  • $\begingroup$ Welcome to MES, +1 $\endgroup$ – k1.M May 4 '15 at 9:24
  • $\begingroup$ It is a consequence of the super-additivity of the geometric mean. $\endgroup$ – Jack D'Aurizio May 4 '15 at 9:34
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Hint $$x_{i}+1\ge 2\sqrt{x_{i}}$$

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  • $\begingroup$ By the AM-GM inequality. Indeed, what I was doing was actually similiar to the proof of AM-GM, but slightly easier. $\endgroup$ – wythagoras May 4 '15 at 9:11
  • $\begingroup$ @wythagoras yes, although you can also think of it as just $(\sqrt{x_i}-1)^2\geq 0$. $\endgroup$ – Stan Liou May 4 '15 at 12:47
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$$ (1+x_1)(1+x_2)\dotsb(1+x_n)=(1+x_1+x_2+\dotsb+x_{n-1}+x_n)+(x_1x_2+x_1x_3+\dotsb+x_{n-1}x_n)+\dotsb+x_1x_2\dotsb x_n $$ Now use AM-GM inequality for the right side of the above equation with $2^n$ terms.

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By Holder $$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\sqrt[n]{x_1x_2...x_n}\right)^n\geq(1+1)^n=2^n.$$ Done!

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0
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Hint: Use the following form of induction:

  • If $P(n)$, then $P(2n)$.
  • If $P(n)$ then $P(n-1)$.
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HINT: Expand the RHS and collect the terms with equal degree. Then use AM-GM on these collections separately. For example, the first such collection would be $$\frac{\sum_{i=1}^nx_i}{\binom n1}\ge(x_1x_2x_3...x_n)^{\frac 1n}\ge 1$$ Hence $$\sum_{i=1}^nx_i\ge \binom n1$$ Add all these results and use the fact that $$\sum_{i=0}^n\binom ni=2^n$$That leads you to the answer.

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We are given that $x_1, x_2, \cdots, x_n$ are positive real numbers and $x_1x_2\cdots x_n\geq 1\forall n\in \mathbb N$.

We have to show $(1+x_1)(1+x_2)\cdots (1+x_n)\geq 2^n$.

For $n=1$ the result is trivial as $x_1\geq 1\Rightarrow (1+x_1)\geq 1+1=2^1.$

Let the result be true for $n=m$ viz $x_1x_2\cdots x_m\geq 1$ implies $$(1+x_1)(1+x_2)\cdots (1+x_m)\geq 2^m$$ Multiplying both sides by the inequality $1+x_{m+1}\geq 2$ we see that $$(1+x_1)\cdots(1+x_m)(1+x_{m+1})\geq 2^{m+1}$$ which shows that the inequality is true for $n=m+1$ whenever it is true for $n=m$.

Induction now completes.

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  • $\begingroup$ Why would $1+x_{m+1} \geq 2$ ? $\endgroup$ – Max Apr 23 '17 at 16:32
  • $\begingroup$ Why would $1+x_{m+1} \geq 2$ ? $\endgroup$ – Max Apr 23 '17 at 16:32
  • $\begingroup$ Because of the case $n=1$ $\endgroup$ – Anjan3 Apr 28 '17 at 15:07
  • $\begingroup$ Who ever said $x_{m+1} \geq 1$ ? Moreover, $x_1...x_m$ need not be $\geq 1$ either, so you can't apply the induction hypothesis like that $\endgroup$ – Max Apr 28 '17 at 17:42
  • $\begingroup$ $x_1x_2\cdots x_n \geqslant 1$ is given already. $\endgroup$ – Anjan3 May 2 '17 at 3:02

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