Inequalities #4

Dec 2015
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Earth
1. Prove that \(\displaystyle n!^{2}\geq n^n \; , n\in \mathbb{N}\).

2. Prove that \(\displaystyle (n+1/n)^n \geq n^2 +n \).
 
Mar 2015
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Universe 2.71828i3.14159
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.
 
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Dec 2015
1,082
169
Earth
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

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Dec 2006
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2. The inequality doesn't hold for $n = \frac32$, so some restriction on $n$ is needed.