# Proving Inequality $\forall x\ge1$

How do I prove that $\forall x\ge1$ $$\left(\frac{x+1}x\right)^x\left(\ln\left\{1+\frac1x\right\}-\frac1{x+1}\right)>0\,?$$

I have tried rearranging this many times but I always end up with a compilicated fraction. Is there a special way I can rearrange this such that it becomes solvable? I'm thinking along the lines of rearranging it such that it satisfies a known thoerem?

We use that, for any $x'\geqslant 0$ $$\tag 1 1-\frac 1{x'+1} \leqslant \log(1+x')$$

Now, plug $x'=x^{-1}$. For the proof of $(1)$, note that for $x''\geqslant 1$, $$1-\frac 1 {x''}=\int_1^{x''}\frac{dt}{t^2}\leqslant \int_1^{x''}\frac{dt}t=\log x''$$

Note that inequality is actually strict whenever $x''>1$, that is $x'>0$.

• To get the desired strict inequality, break into cases: use the fact that $$\left(\frac1t\right)^2<\frac1t$$ when $t>1$ to show that the inequality is strict when $x>1$, and just directly compute it when $x=1$. – Karl Kronenfeld Aug 26 '13 at 4:14
• @KarlKronenfeld Added. – Pedro Tamaroff Aug 26 '13 at 4:16

Hint: The $\left(\frac{x + 1}{x}\right)^x$ term is always positive for $x \ge 1$, so it's irrelevant. So it suffices to show that

$$\ln\left(1 + \frac{1}{x}\right) > \frac{1}{1 + x}$$

This is a much simpler inequality to consider.