Answer to Question #88221 in Algebra for O. P. Kaith

Question

Prove that: 2^n > 1+n√2^(n-1) for all n>2 using the concept of inequalities only.

Answer 1

Inequality Between Means (AM-GM inequality)

Let $a_1, a_2,...,a_n,$ be positive real numbers. Then

$${a_1+a_2+...+a_n \over n}\geq \sqrt[n]{a_1a_2...a_n}$$

Equality occurs if and only if $a_1=a_2=...=a_n$ or $n=1.$

Consider the geometric sequence $1,2,4,8,...$

$$b_1=1, q=2,$$ $$S_n=b_1+b_2+...+b_n=\displaystyle\sum_{i=1}^n b_i=1+2+...+2^{n-1}=$$ $$=b_1({1-q^n \over 1-q})=1({1-2^n \over 1-2})=2^n-1$$

$$\sqrt[n]{b_1b_2...b_n}=\sqrt[n]{1\cdotp2\cdotp4\cdotp...\cdotp2^{n-1}}$$

$$1+2+3+...(n-1)={1+n-1 \over 2}\cdotp (n-1)={n(n-1) \over 2}$$

For $n\geq 2$ we have

$${1+2+...+2^{n-1} \over n}> \sqrt[n]{1\cdotp2\cdotp4\cdotp...\cdotp2^{n-1}}$$

$${2^n-1 \over n}> 2^{{n(n-1) \over {2n}}}$$

$${2^n-1 \over n}> 2^{{n-1 \over {2}}}$$

$$2^n>1+(\sqrt{2})^{n-1} ,n\geq2$$