Show $(1+x_1)(1+x_2)...(1+x_n)\geq2^n$ given $x_1x_2...x_n=1$ and that all $x_i $ are positive reals.
I think simple AM-GM-HM must work, but I am missing on something trivial.
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Michael Rozenberg
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Soham
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Please include the question in the body of the post, not only in its title. // Got anything from an answer below? – Did May 06 '17 at 07:29
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See also Inequality in Algebra: $1 \leq x_1 x_2 \cdots x_n$ implies that $2^{n} \leq (1 + x_1)(1+x_2) \cdots (1 + x_n).$ and Let $a_1,a_2,\dots,a_n$ $\in$ $\mathbb{R}^+$ and $a_1\cdot a_2\cdots a_n=1$, prove that $(1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n$. Found using Approach0. – Martin Sleziak Aug 19 '17 at 19:04
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By AM-GM inequality,
$$(1+x_i) \geq 2 \sqrt x_i$$
Taking product from $i=1$ to $n$, we get the desired result.
MathGod
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1I like how this avoids using the more general AM-GM that some students aren't familiar with and keeps it as elementary as can be. – Chris Apr 13 '17 at 20:18
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Not substantially different form other answers, but perhaps an interesting manipulation: multiply the LHS by $$1=1^{-1/2}=\prod_{k=1}^{n}x_k^{-1/2}$$ obtaining $$\prod_{j=1}^k(1+x_j)=\prod_{j=1}\left(\frac1{x_j^{1/2}}+x_j^{1/2}\right)$$
And then use the fact that (for the reason you prefer) $$\min\left\{ \frac1x+x\,:\,x>0\right\}=2$$
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The brute force approach is:
$$(1+x_1)(1+x_2)\cdots(1+x_n)=\sum_{S\subseteq \{1,\dots,n\}}\prod_{i\in S}x_i$$
But by AM/GM, since there are $2^n$ subsets of $\{1,\dots,n\}$, you get:
$$\frac{1}{2^n}\sum_{S\subseteq \{1,\dots,n\}}\prod_{i\in S}x_i \geq \sqrt[2^n]{x_1^{2^{n-1}}\cdots x_n^{2^{n-1}}}=1$$
This is because:
$$\prod_{S\subseteq \{1,\dots,n\}}\prod_{i\in S} x_i=x_1^{2^{n-1}}\cdots x_n^{2^{n-1}}$$
Thomas Andrews
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By Holder $$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\sqrt[n]{x_1x_2...x_n}\right)^n=2^n$$
Michael Rozenberg
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