How can I prove through mathematical induction that $(n!)^{(1/n)}\le(n+1)/2$?

Question

$$(n!)^{(\frac1n)}\le\frac{(n+1)}2$$

I am unable to prove this proposition through mathematical induction. Can someone help me with it?

Answer 1

By AM-GM we get $$\sqrt[n]{1\cdot 2\cdot 3\cdots n}\le \frac{1+2+3+\cdots +n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$$

Answer 2

Start from:

$$2^nn!\le (n+1)^n$$

Multiply left and right by $2(n+1)$:

$$2^{n+1}(n+1)!\le 2(n+1)^{n+1}\le (n+2)^{n+1}$$

And:

$$\left(1+\frac1n\right)^{n+1}\to e\gt 2$$