Prove that $\frac{x}{y} \leq \ln(y) - \ln(y - x)$

Question

I am supposed to prove that $\frac{x}{y} \leq \ln(y) - \ln(y-x)$ for $x, y \in \mathbb{R}$ with $0 \leq x < y$.

So I've been thinking about going with a case distinction. First is the obvious case for $x=0$. Then I thought about doing more case distinctions where I differ whether $|y - x|$ is smaller, equals or larger to 1. But already in th easiest case for $|y-x| = 1$ I pretty much get stuck. I have no idea how I'm supposed to get the logarithm into the inequality.

So I'd be more than glad about a tiny hint here.

Answer 1

We know $e^t \geq t + 1 $ for all $t$. In particular, $e^{-t} \leq \frac{1}{1+t}$.

Put $t = - \frac{x}{y}$. Then,

$$ e^{\frac{x}{y}} \leq \frac{1}{1 - \frac{x}{y}} \implies \frac{x}{y} \leq \ln \left( \frac{y}{y-x} \right)$$

as we wanted to show.

Answer 2

Hint:

Set $x=ty,\quad 0\le t<1$. You have to proove $$t\le \ln y-\ln(y-x)=-\ln(1-t).$$ Use the Taylor's series expansion of $\ln(1-t)$.

Answer 3

Consider the function $t\in [y-x,y]\to \ln t.$ Because of the mean value theorem for derivatives there exists $c\in (y-x,y)$ such that

$$\ln y-\ln(y-x)=\frac{d\ln}{dx}|_c (y-(y-x))=\frac{x}{c}.$$ Now, $x/c>x/y$ since $c<y.$