Show $\left(1-\frac x k\right)^k<\left(1-\frac {x}{ k+1}\right)^{k+1}$

Question

Show that $$\left(1-\frac{x}{k}\right)^k<\left(1-\frac{x}{k+1}\right)^{k+1}$$ for $x>0$, and $k \ge 1$, where $k$ is a whole number.

Is it possible to prove this? I can easily prove algebraically for $k=1$ case, but I am wondering if this is true in general.

Answer 1

Hint:

$\frac{d}{dk}\left( e^{k\ln\left(1-\frac{x}{k}\right)} \right)$ is positive for all positive $k \ge 1$ [$k \in \mathbb{R}; k-1 >0; k > x$]

Answer 2

Consider $f(x) = k\ln(k-x)- k\ln k - (k+1)\ln(k+1-x) + (k+1)\ln(k+1), k>x > 0, k \ge 1\implies f'(x) = -\dfrac{k}{k-x}+\dfrac{k+1}{k+1-x}= - \dfrac{x}{(k-x)(k+1-x)}<0\implies f(x) < f(0)=0$ and this implies the result. For if $x = k$, the result is clearly true, and lastly if $x > k$ and if $k$ is odd, the $LHS < RHS$ and it is true again. If $k$ is even, can you continue to finish it off?

Answer 3

Consider numbers $1$, $1-\dfrac{x}{k}$ ($k$-times) and apply $AM-GM$ then $$\dfrac{1+(1-\dfrac{x}{k})+(1-\dfrac{x}{k})+(1-\dfrac{x}{k})+\cdots+(1-\dfrac{x}{k})}{k+1}\geq\sqrt[k+1]{\left(1-\dfrac{x}{k}\right)^k}$$ Simplify and find your result.

Answer 4

Want $\left(1-\frac{x}{k}\right)^k \lt\left(1-\frac{x}{k+1}\right)^{k+1} $.

We must have $0 < x < k$.

Let $f(k) =k\ln(1-x/k) $.

$\begin{array}\\ f'(k) &=\ln(1-x/k)+k\dfrac{(1-x/k)'}{1-x/k}\\ &=\ln(1-x/k)+k\dfrac{x/k^2}{1-x/k}\\ &=\ln(1-x/k)+\dfrac{x/k}{1-x/k}\\ &=\ln(1-z)+\dfrac{z}{1-z} \qquad z = x/k \text{ so } 0 < z < 1\\ &=-\sum_{n=1}^{\infty}\dfrac{z^n}{n}+\sum_{n=1}^{\infty}z^n\\ &=\sum_{n=1}^{\infty}z^n(1-\frac1{n})\\ &=\sum_{n=2}^{\infty}z^n(1-\frac1{n})\\ &> 0\\ \end{array} $

Therefore $f(k)$ is an increasing function of $k$.

Answer 5

It's true if $k \geqslant x$ and it follows from Bernoulli's inequality

$$\left(1 - \frac{x}{k+1} \right)^{(k+1)/k} > 1 - \frac{k+1}{k}\frac{x}{k+1} = 1 - \frac{x}{k} \\ \implies \left(1 - \frac{x}{k+1} \right)^{k+1} > \left(1 - \frac{x}{k} \right)^{k}$$