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?

See also: https://math.stackexchange.com/q/79109/42969, https://math.stackexchange.com/q/1736503/42969. — Martin R, Nov 24 '19 at 14:09

score 4 · Answer 1 · answered Nov 24 '19 at 13:30

4

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

answered Nov 24 '19 at 13:30

This is so beautiful. – Matha Mota Nov 24 '19 at 13:41
Thank you for the flowers! – Dr. Sonnhard Graubner Nov 24 '19 at 13:42
You're welcome. – Matha Mota Nov 24 '19 at 13:42
3

The question is a duplicate, and you gave an identical answer here: https://math.stackexchange.com/a/2202255/42969. – Martin R Nov 24 '19 at 14:09

score 0 · Answer 2 · answered Nov 24 '19 at 13:32

0

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:

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

answered Nov 24 '19 at 13:32

JMP

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