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Please give me the proof for the following: Let $a_1,\,a_2,\,\dots\,a_n$ be $n$ positive real numbers whose product is equal to $1$. Prove that $$a_1+\cdots+a_n \geq n$$ and that the equality sign holds only if every $a_k$ is equal to $1$.

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  • $\begingroup$ Use well ordering principle en.wikipedia.org/wiki/Well-ordering_principle $\endgroup$ – ys wong Nov 30 '15 at 14:24
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    $\begingroup$ @yswong How is this related to that principle? $\endgroup$ – Kibble Nov 30 '15 at 14:38
  • $\begingroup$ @yswong This doesn't look like an easy proof by induction. $\endgroup$ – BrianO Nov 30 '15 at 14:54
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This follows from the AM-GM inequality. The shortest proof makes use of the Jensen Inequality.

The function $f(x)= \ln x$ is concave since $f''(x)=-\frac{1}{x^2}<0$.

Now by Jensen Inequality, we have $$f(\frac{\sum_{i=1}^n a_i}{n}) \ge \frac{1}{n} \sum_{i=1}^n f(a_i)$$

Taking the exponential function, we have $$\frac{1}{n} \sum_{i=1}^n a_i \ge \sqrt[n]{\prod_{i=1}^n a_i}$$

Since $\prod_{i=1}^n a_i = 1$, we have $$\sum_{i=1}^n a_i \ge n$$ as desired.

The equality holds if $a_i=1$ for all $1 \le i\le n$.

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If $n=1$, then $a_1=1$ which is true. Suppose that, for $n=k$ and $a_1a_2\cdots a_k=1$, $$ a_1+a_2+\cdots a_k\ge k.$$ For $n=k+1$, since $a_1a_2\cdots a_ka_{k+1}=1$, we have that $a_1a_2\cdots a_{k-1}(a_ka_{k+1})=1$ implies $$ a_1+a_2+\cdots a_{k-1}+a_ka_{k+1}\ge k.$$ Note that we can choose $a_k<1$ and $a_{k+1}>1$ since if $a_1=a_2=\cdots=a_k=a_{k+1}=1$, the inequality is true and otherwise we have some $a_i<1$ and some $a_{j}>1$ and then switch $a_i$ with $a_k$ and $a_j$ with $a_{k+1}$. Thus $$ a_1+a_2+\cdots a_{k-1}+a_k+a_{k+1}\ge k+a_k+a_{k+1}-a_ka_{k+1}=k+1+(1-a_k)(a_{k+1}-1)> k+1.$$

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