How to show $\ln{(\frac{x+1}{x})}\geq \frac{1}{x+1}$?

Question

I used mathematica to check the values of $\ln{(\frac{x+1}{x})}- \frac{1}{x+1}$ are positive for any $x\geq 0$.

How do I analytically show the statement is true for all $x\geq 0$?

Any idea?

Answer 1

By the definition of the natural logarithm $$\ln\left({\frac{x+1}{x}}\right):=\int_{1}^{1+\frac1x}\frac1tdt\ge(1+\frac1x-1)\cdot \frac{1}{1+\frac1x}= \frac1{x+1}$$ where the inequality is actually strict.

Answer 2

It should be $x>0.$

We need to show that $$\ln\frac{x}{x+1}\leq-\frac{1}{x+1}$$ or $$\ln\left(1-\frac{1}{1+x}\right)\leq-\frac{1}{1+x}.$$ It's enough to prove that $$\ln(1+x)\leq x$$ for all $x>-1.$

Can you end it now?

Answer 3

Let $f(t)=\ln (1+t)-\frac t {1+t}$. Then $f(0)=0$ and $f'(t)=\frac 1 {1+t}-\frac 1 {(1+t)^{2}}>0$ for all $t>0$. Hence $f(t) >0$ for all $t \in (0,\infty)$. If you put $t =\frac 1 x$ in this you will get the result you want.