Question

Prove that $\frac{1}{m}\log (1+a^m)<\frac{1}{n}\log (1+a^n)$, if $m>n$.

Answer 1

${(1+a^m)}^n={(a^m+1)}^n$

Expansion of ${(a^m+1)}^n{=}^n{C}_0{\left(a^m\right)}^n{+}^n{C}_1{\left(a^m\right)}^{n-1}+...+1$

Taking first two term only we get,

${}^n{C}_0{\left(a^m\right)}^n{+}^n{C}_1{\left(a^m\right)}^{n-1}=a^{mn}+n{a}^{mn-n}=a^{mn}(1+n{a}^{-m})$

${(1+a^n)}^m={(a^n+1)}^m$

Expansion of ${(a^n+1)}^m={\left(a^n\right)}^m{+}_1^m\text{C}{\left(a^n\right)}^{m-1}+...+1$

Taking first two term only we get,

${}^m{C}_0{\left(a^n\right)}^m{+}^1{C}_m{\left(a^n\right)}^{m-1}=a^{mn}+m{a}^{mn-n}=a^{mn}(1+m{a}^{-n})$

Given $m>n$

$a^m>a^n$

$a^{-m}<a^{-n}$

$n{a}^{-m}<m{a}^{-n}$

$1+n{a}^{-m}<1+m{a}^{-n}$

$a^{mn}(1+n{a}^{-m})<a^{mn}(1+m{a}^{-n})$

${(1+a^m)}^n<{(1+a^n)}^m$

Takinf $\log$ both side

$\log {(1+a^m)}^n<\log {(1+a^n)}^m$

$n\log (1+a^m)<m\log (1+a^n)$

$\frac{1}{m}\log (1+a^m)<\frac{1}{n}\log (1+a^n)$