Prove by induction $n^{1/n} ≤ \frac{n+1}{2}$

Question

The problem

Prove by induction:

$n^{1/n} ≤ \frac{n+1}{2}$

Attempt at solution

I started off with the usual steps for an MI problem.

We start with the $P_1$ case:

for $P_1$, LHS = 1 and RHS = 1 $\implies$ $P_1$ is true.

Assume that $P_k$ is true for some $k \in \mathbb{N}^+$:

$k^{1/k} ≤ \frac{k+1}{2}$

We need to show that $P_{k+1}$ is true:

i.e $(k+1)^{1/(k+1)} ≤ \frac{k+2}{2}$

I rearranged $P_k$ to give $2k^{1/k} ≤ k+1$

I then thought to substitute this back into the LHS of $P_{k+1}$:

$(2k^{1/k})^{1/(k+1)}\leq \frac{k+2}{2}$

I then tried manipulating with rules of indices, but to no avail. Additionally I don't even know whether its valid to substitute something into the inequality like I just did.

Answer 1

Hint: Prove by induction that $n\le\left(\frac{n+1}2\right)^n$

Actually, it is abit absurd to try this by induction once you observe that $\left(\frac{n+1}2\right)^n\ge\left(\frac{n+1}2\right)^2>n$ for all but the smallest $n$