Show that (a_1+1)(a_2+1)\cdots(a_n+1)>=2^n
Solution
We have using AM GM \frac{a_k+1}{2} >= \sqrt{a_k} or a_k + 1>=2\sqrt{a_k}
multiplying over k from 1 to n we have
\prod_{k=1}^n(a_k+1) >= 2^n \sqrt{\prod_{k=1}^na_k} = 2^n
or \prod_{k=1}^n(a_k+1) >= 2^n
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