1/x+1 less than or equal to ln(x+1)/x
using definition of ln(x) as intergral of 1 to x of 1/t dt
show that for every real number we have x such that
1/x+1 less than or equal to ln(x+1)/x
using definition of ln(x) as intergral of 1 to x of 1/t dt
show that for every real number we have x such that
By definition: if $\displaystyle x>0$ then $\displaystyle \ln (x) = \int_1^x {\frac{{dt}}{t}} $.
So $\displaystyle \left( {\ln (x)} \right)^\prime = \frac{1}{x}$ which means we can use the mean value theorem.
If $\displaystyle 0<a<b$ then $\displaystyle \left( {\exists c \in (a,b)} \right)\left[ {\frac{1}{c} = \frac{{\ln (b) - \ln (a)}}{{b - a}}} \right]$
But $\displaystyle a<c<b$ implies $\displaystyle \frac{1}{b}<\frac{1}{c}<\frac{1}{a}$
That gives $\displaystyle \frac{1}{b}<\frac{{\ln (b) - \ln (a)}}{{b - a}}}<\frac{1}{a}$.
Let $\displaystyle a=1~\&~b=x+1$ and we have it.