Prove the inequality involving exponential function in form of $\exp( \frac{1}{x} )$

Question

For $\nu > 0$, $0 < x \leq \nu $, and a positive integer $S$, (we think) following an inequality always holds

$1- \left( \frac{1}{x+1} \right)^S \geq \exp \left( -\frac{1}{Sx} \right) $

Does anyone help us for proving it?

Answer 1

When $S=1$, we have $$ \exp(1/x)\geq 1+1/x=\frac{x+1}{x}\iff\exp(-1/x)\leq\frac{x}{1+x}=1-\frac{1}{x+1}. $$ Suppose that the claim is true for some $S\geq 1$. Then, we have $$ \exp(S/x)\geq[1-(1/(x+1))^S]^{-1}\quad\text{and}\quad\exp(1/x)\geq[1-1/(x+1)]^{-1}. $$ Multiplying these inequalities, we have $$ \exp[(S+1)/x]\geq [1-(1/(x+1))^S]^{-1}[1-1/(x+1)]^{-1} $$ So it suffices to show $$ \left(1-\frac{1}{(x+1)^S}\right)\left(1-\frac{1}{x+1}\right)\leq\left(1-\frac{1}{(x+1)^{S+1}}\right).\tag{$*$} $$ But this is true: for ($*$), $$ \text{RHS}-\text{LHS}=\frac{1}{(x+1)^{S+1}}\left[(x+1)^S-1+x\right]>0. $$ (For the term in the square brackets above, one can for example use the Bernoulli's inequality $(x+1)^S-1+x\geq xS+1-1+x=x(S+1)>0$.)