3
$\begingroup$

I conjecture the following inequality is true $$\ln x \le (x - 1)\ln\frac{x}{x-1}$$ for all $x > 1$, but I cannot give a proof. I will appreciate if someone can provide one.

  • 5
    Lord Wolfram says, the solution of your inequality is $1, and not $x > 1$. So, your conjecture is disproved!! http://www.wolframalpha.com/input/?i=ln+x+%3C%3D+%28x-1%29+ln+%28x%2F%28x-1%29%292011-12-24
  • 2
    Actually, for $x \geq 2$, ln$x \geq (x-1)$ln$(x/x-1)$ is true. http://www.wolframalpha.com/input/?i=ln+x+%3E%3D+%28x-1%29+ln+%28x%2F%28x-1%29%292011-12-24
  • 0
    Could you explain how you conjecture this?2011-12-24
  • 0
    Make the substitution $y = x - 1$ and exponentiate to get the conjectured inequality of $1+y \leq (1+1/y)^y$ for all $y > 0$. Since the right-hand side converges to $e$ as $y \to \infty$, then the conjecture is obviously false.2011-12-24

3 Answers 3