2

I've stumbled over the following inequality while reading a book:

$$e^{x/(k+1)} \leq 1 + x/k \quad\text{for}\quad\text{$0 \leq x \leq 1$ and $k \in \mathbb N$}$$

The inequality is given without proof and without references. I think it is non-trivial and would like to understand, where it is coming from. I have searched for popular inequalities concerning the exponential function but haven't found this one.

Does this inequality have a name? (probably not)

Does anyone know a source where I can read up on this inequality?

If not, does anyone know how to prove it?

kostrykin
  • 475

3 Answers3

2

Taking the logarithm on both side we have to show for $0\leq x\leq 1$ and $k$ a stricly positive natural number :

$$\frac{x}{k+1}-\ln\left(1+\frac{x}{k}\right)\leq 0$$

Define :

$$f(x)=\frac{x}{k+1}-\ln\left(1+\frac{x}{k}\right)$$

Differentiate with respect to $x$ we get :

$$f'(x)=(x-1)/(somethingpositive)$$

So $f$ is decreasing on $x\in[0,1]$ .

But : $f(0)=0$ wich yields the inequality.

Is it ok for you ?

Miss and Mister cassoulet char
  • 3,511
2

\begin{multline*} \exp\left(\frac{x}{k+1}\right) = 1 + \frac{x}{k+1} + \frac{x^2}{2!}\left(\frac1{k+1}\right)^2 + \frac{x^3}{3!}\left(\frac1{k+1}\right)^3 + \cdots \\ \leqslant 1 + \frac{x}{k+1}\left(1 + \frac1{k+1} + \left(\frac1{k+1}\right)^2 + \cdots\right) = 1 + \frac{x}{k}. \end{multline*}

Calum Gilhooley
  • 12,174
2

Your inequality is also a consequence of the “well-known” inequality $1+x \le e^x$ (see here for various proofs).

Using that estimate we get $$ e^{x/(k+1)} = \frac{1}{e^{-x/(k+1)}} \le \frac{1}{1-x/(k+1)} = \frac{k+1}{k+1-x} \, , $$ and the desired inequality follows, since $$ \frac{k+1}{k+1-x} - \left( 1 + \frac xk \right) = -\frac{x(1-x)}{k(k+1-x)} \le 0 \, . $$

Martin R
  • 113,040