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I want to know that can we have the following inequality for $x>0$, $y>0$?

$$\log(1+x/y) \leq x/y$$

egreg
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Ramin
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  • See https://math.stackexchange.com/q/1376462/42969 or https://math.stackexchange.com/q/652581/42969 or https://math.stackexchange.com/q/324345/42969 ... all of them with links to more duplicate questions – Martin R Feb 22 '18 at 16:22

4 Answers4

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Yes.

It follows from $$e^t\ge 1+t $$ for all $t\in\Bbb R$ (with equality iff $t=0$).

Hagen von Eitzen
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Yes you can.

Write $t=x/y>0$ and $f(t) = t-\log(1+t)$, then $f(0) = 0$ and $$f'(t) = 1-{1\over 1+t} = {t\over 1+t} >0$$

So $f$ is increasing, so for every $t>0$ we have $f(t)>0$.

nonuser
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Let $f(t)=t-\ln(1+t),$ where $t>0$.

Thus, $$f'(t)=1-\frac{1}{1+t}=\frac{t}{1+t}>0,$$ which says $$f(t)>f(0)=0.$$

For $t=\frac{x}{y}$ we obtain: $$\frac{x}{y}>\ln\left(1+\frac{x}{y}\right).$$

Michael Rozenberg
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Know that for $x>y$, the inequality holds and is strict. For $x<y$, let's expand $\log(1+\frac{x}{y})$ around $x\approx 0$,

$ \log(1+\frac{x}{y}) \approx \frac{x}{y} - \frac{x^2}{2y^2} + \frac{x^3}{3y^3} - \frac{x^4}{4y^4} + \dots $

Now, since $\frac{x^2}{2y^2} > \frac{x^3}{3y^3}$, we are adding a negative number to $\frac{x}{y}$ in the above expression. So, $\exists$ some $\Delta>0$ such that

$ \log(1+\frac{x}{y}) \leq \frac{x}{y} - \Delta. $

unobservable_node
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