# Prove that $\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}$ when $x > 1$ [closed]

We want to prove that

$$\forall x>1, \,\,\,\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}.$$

I tried to look up inequalities involving $$\log$$ that might be useful, but I couldn't finagle any of them to fit this particular situation.

• what is the base of log? Feb 4, 2021 at 2:32
• Applying $10^x$ both sides? Feb 4, 2021 at 2:33
• It's thoroughly incorrect. ,$x=0.5$ is a counterexample. Feb 4, 2021 at 2:35
• @Wesley It's false. Did you mean $x>1$? Feb 4, 2021 at 2:38
• @jjagmath Oh right, sorry. I meant $x > 1$ Feb 4, 2021 at 3:08

By strict concavity of $$f(x) = \ln(x)$$, we have that $$f(y) < f(x) + f'(x)(y-x) \implies \ln(y) < \ln(x) + \frac{y-x}{x},\forall x,y\in\mathbb{R}_{>0}.$$ Setting $$y=1+x$$ and letting $$x$$ be itself, we then have for $$x>1$$: $$\ln(1 + \frac{1}{x}) = \ln(\frac{x+1}{x}) = \ln(1+x) - \ln(x) < \frac{1+x-x}{x} = \frac{1}{x} < \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}$$ where the latter equality follows from partial fraction expansion.

Note that, for $$x>1$$,

\begin{align} f(x)&=\frac{2x-1}{x^2-x}-\ln\left(1+\frac{1}{x} \right)\\ &=-\int_x^\infty f’(t)dt=\int_x^\infty \frac{(t-1)^3+5(t-\frac12)^2+\frac34}{(t^2-t)^2(t+1)}dt>0 \end{align}

• Huh? Where? What? How? Is that integral some identity?
– sato
Feb 4, 2021 at 3:50
• @Mastermind817 - the integrand is just $-f’(t)>0$ Feb 4, 2021 at 4:00

Consider that you look for the minimum value of function$$f(x)= \frac{2x-1}{x^2-x}-\ln\left(1+\frac{1}{x} \right)\qquad \text{with} \qquad x>1$$ Its first derivative is $$f'(x)=\frac{-x^3-2 x^2+2 x-1}{(x-1)^2\, x^2\, (x+1)}$$ the denominator is positive $$\forall x >1$$. The numerator shows only one real root $$(\Delta=-83)$$ which is $$x=-\frac{2}{3} \left(1+\sqrt{10} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{79}{20 \sqrt{10}}\right)\right)\right) \sim -2.83118$$ which is not in the acceptable range. So $$f'(x) <0 \quad \forall x >1$$ and $$f(x)$$ is a decreasing function which starts from $$+\infty$$.

Around $$x=1$$, by Taylor $$f(x)=\frac{1}{x-1}+(1-\log (2))-\frac{x-1}{2}+O\left((x-1)^2\right)$$

Now, for large values of $$x$$ $$f(x)=\frac 1x +\frac 3{2x^2}+\sum_{n=3}^\infty \frac{n+(-1)^n}{n\, x^n}$$ and all coefficients are positive.

Hence, $$f(x) >0 \to 0^+$$ and the inequality holds.