Math Help - Show 1/x+1 less than or equal to ln(x+1)/x

1. Show 1/x+1 less than or equal to ln(x+1)/x

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

Originally Posted by sbacon1991
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 $x>0$ then $\ln (x) = \int_1^x {\frac{{dt}}{t}}$.

So $\left( {\ln (x)} \right)^\prime = \frac{1}{x}$ which means we can use the mean value theorem.

If $0 then $\left( {\exists c \in (a,b)} \right)\left[ {\frac{1}{c} = \frac{{\ln (b) - \ln (a)}}{{b - a}}} \right]$

But $a implies $\frac{1}{b}<\frac{1}{c}<\frac{1}{a}$

That gives $\frac{1}{b}<\frac{{\ln (b) - \ln (a)}}{{b - a}}}<\frac{1}{a}$.

Let $a=1~\&~b=x+1$ and we have it.