Prove Inequality using Induction

Question

I am stuck on the last step where I need to prove: $$k^\frac{1}{k} > (k+1)^\frac{1}{k+1}$$

Please give me methods only involving induction. Thank you.

Answer 1

Raise both sides to $k(k+1)$ we get:

$k^{k+1} > (k+1)^k$

To prove: $ k > (1 + \frac{1}{k})^k$

Assume true for $k = n$; and now consider the case $k = n+1$

Then $n + 1 > (1 + \frac{1}{n})^n + 1$

multiplying both sides with $(1 + \frac{1}{n})$ we get

$(n +1)(1 + \frac{1}{n}) = n + 1 + 1 + \frac{1}{n} > (1 + \frac{1}{n})^{n+1} + 1 + \frac{1}{n} > (1 + \frac{1}{n+1})^{n+1} + 1 + \frac{1}{n}$

$\implies n + 1 > (1 + \frac{1}{n+1})^{n+1}$

Answer 2

You can get $\log$ from each side of the inequality: $$\frac{\log k}{k} > \frac{\log (k+1)}{k+1}$$

Then, we want to show that $f(x) = \frac{\log x}{x}$ is a decreasing function. As $f'(x) = \frac{1-\log x}{x^2}$ is less than zero for $x > 1$, Hence the proof is completed.

Answer 3

I give a proof by induction using only Bernoulli's inequality here:

Is this proof that $\lim_{n \to \infty} (1+1/n)^n$ exists (1) new, (2) interesting?