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.
Inequality for logarithms
3
$\begingroup$
analysis
inequality
logarithms
-
5Lord 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%29 – 2011-12-24 -
2Actually, 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%29 – 2011-12-24
-
0Could you explain how you conjecture this? – 2011-12-24
-
0Make 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