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.

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

3 Answers 3


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}

  • 1
    $\begingroup$ Huh? Where? What? How? Is that integral some identity? $\endgroup$
    – sato
    Feb 4, 2021 at 3:50
  • $\begingroup$ @Mastermind817 - the integrand is just $-f’(t)>0$ $\endgroup$
    – Quanto
    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.


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