Is it possible to prove this? $\ln(\frac{x}{x-1}) < \frac{100}{x} $ for $ x > 1$

Question

$-\ln(1-(\frac{1}{x})) < \frac{100}{x} $ for $ x > 1$ is what I want to prove. I pulled a negative sign out and I got $\ln(\frac{x}{(x-1)}) < \frac{100}{x} $ for $ x > 1$. How do I continue with this proof? Or is it actually possible to prove this

Edit : I want this proof Algebrically, Calculus is allowed

Answer 1

$ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$

$-ln \left(\frac{x-1}{x}\right)<\frac{100}{x}$

$-ln \left(1-\frac{1}{x}\right)<\frac{100}{x}$

$ln \left(1-\frac{1}{x}\right)^x>-100$

$\left(1-\frac{1}{x}\right)^x>e^{-100}$

$\left(1-\frac{1}{x}\right)^x>\frac{1}{e^{100}}$

I think that $\frac{1}{e^{100}}$ is almost 0 and LHS is ofcourse $>0$ for all $x>1$

So this seems pretty reasonable.

Edit: It suddenly seems not strictly valid for all $x>1$ , but not sure. Edit:2 This is not true for only

$x\in(1,root\;of\;(x-1)-\frac{x}{e^{100/x}}=0)$

Since, the root lies very close to $1$ , it is approximately valid. And I think, this is good enough for algebraic solution.

Answer 2

It is not true for $x=1+e^{-100}$. We then have $$ \log\frac{x}{x-1} = \log(x)+100 > 100 $$ but $$ \frac{100}{x} < 100 $$ because $x>1$.

Answer 3

Solution to \begin{align} \ln\left(\frac{x}{x-1}\right) = \frac{100}{x} \end{align} in terms of Lambert W function is \begin{align} x&=\frac{100}{W(-100\exp(-100))+100} \\ &\approx 1.000000000000000000000000000000000000000000037200759760 \end{align}

Btw, WolframAlpha gives the answer to the inequality as

Solution over the reals

Answer 4

As you almost wrote, making $x=\frac 1y$, the problem is to show that, for $y<1$, $$f(y)=100 y+\log(1-y)>0$$ (the function being undefined at $y=1$). We have $$f'(y)=100-\frac{1}{1-y}$$ which is positive as long as $y<\frac{99}{100}$ and negative for $y>\frac{99}{100}$. On the other hand $$f''(y)=-\frac{1}{(1-y)^2}$$ is always negative. So $f(y)$ increases up to a maximum value equal to $99-\log (100) >0$ and starts decreasing very fast and it exists a value of $y$ slightly below $1$ which makes $f(y)=1$ and negative for larger values. When $y=1-\epsilon$ , $f(y)= 100-100\epsilon+\log(\epsilon)$

So, as Henning Makholm answered, this is not true for all $x>1$.

Answer 5

Required to prove $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$

ie to prove $f(x) = {{100} \over x} - ln\left( {{x \over {x - 1}}} \right)>0$

${d \over {dx}}\left({{100} \over x} - ln\left( {{x \over {x - 1}}} \right)\right) = $

=......

=${{ - 101x - 100} \over {\mathop x\nolimits^2 (x + 1)}}$

$f'(x)$ is negative for all $x > 1$

Meaning always decreasing.

Easy to show $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$

So The function is always decreasing towards 0

implies $f(x) > 0$ for $x > 1$

Implies $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$