If $n$ be a positive integer greater than $2$ , show that
$2^{n} > 1 + n \sqrt{2^{n - 1}}$ .

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Solution

Using the relation A.M. > G.M. on the numbers $1, 2, 2^{2}, 2^{3} . . . . . . . . . . 2^{n - 1}$ we have
$\frac{1 + 2 + 2^{2} + 2^{3} . . . . . . . . . . 2^{n - 1}}{n} > {(1.2. 2^{2} . 2^{3} . . . . . . . . . . 2^{n - 1})}^{\frac{1}{n}}$
Equality does not hold as all the numbers are not equal.
$\Rightarrow \frac{2^{n} - 1}{2 - 1} > n {⎛ ⎜ ⎜ ⎝ 2^{\frac{n (n - 1)}{2}} ⎞ ⎟ ⎟ ⎠}^{\frac{1}{n}} \Rightarrow 2^{n} - 1 > n (2^{\frac{(n - 1)}{2}}) \Rightarrow 2^{n} > 1 + n (2^{\frac{(n - 1)}{2}})$

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