# Inequalities #4

#### idontknow

1. Prove that $$\displaystyle n!^{2}\geq n^n \; , n\in \mathbb{N}$$.

2. Prove that $$\displaystyle (n+1/n)^n \geq n^2 +n$$.

#### skipjack

Forum Staff
Shouldn't you have $n \in \mathbb{N}$ specified for the second problem as well as the first?

• idontknow and topsquark

#### tahirimanov19

1. Prove that $$\displaystyle n!^{2}\geq n^n \; , n\in \mathbb{N}$$.

2. Prove that $$\displaystyle (n+1/n)^n \geq n^2 +n$$.
1. For n=1 and 2 it's equal and for n=3 left side is greater than right side. Let's assume for n=k it is also true.
$$(k+1)!^2=k!^2 \cdot (k+1)^2 \ge k^k \cdot (k+1)^2$$
Now, prove $k^k \ge (k+1)^{k-1}$. It is easy. :spin:

2. I don't think it is only for $n \in \mathbb{N}$. If it was then proving $n^n+A>n^2+n$ would be too easy.

• 2 people

#### idontknow

I mean without induction .

#### idontknow

I mean without induction .
$$\displaystyle 1\geq n^n n!^{-2} =f(n)$$ and $$\displaystyle \dfrac{f(n+1)}{f(n)}=\dfrac{(1+1/n ) ^n }{1+n } \leq 1$$ , the function is decreasing.

The range of the function is $$\displaystyle f(n\rightarrow \infty ) =0 < f(n) \leq f(1)=1\leq 1$$ , which proves the inequality.

#### skipjack

Forum Staff
2. The inequality doesn't hold for $n = \frac32$, so some restriction on $n$ is needed.

#### idontknow

2. The inequality doesn't hold for $n = \frac32$, so some restriction on $n$ is needed.
My mistake on typing. Let's skip this old thread.