Show that for positive real numbers 
such that 
, then
(1+a_2)\cdots(1+a_n)\ge2^n)
.
Consider the average of the positive numbers x and y: 
. The AM-GM inequality tells us that
and so 
for any positive numbers x and y. Letting 
and 
, we then see that \ge2\sqrt{a_i})
for all i=1,2,…,n. Multiplying these n inequalities together:
\ge\prod_{i=1}^{n}2\sqrt{a_i}=2^n\sqrt{\prod_{i=1}^{n}a_i}=2^n)
,
where the last is because 
.
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Tags: AM-GM Inequality, Math, Monday Math
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