Prove $\binom{2n}{n}\geq\frac{4^n}{2n}$

Question

I want to prove that$$\binom{2n}{n} \geq \frac{4^n }{ 2n}$$

I tried to solve with Stirling formula and got to $$\binom{2n}{n} \geq C*\frac{4^n}{ 2n}.$$ I'm not sure how to continue since $C$ could be between $0$ and $1$. Would greatly appreciate any help.

Possible duplicate of A classical inequality involving the central binomial coefficient (see also https://math.stackexchange.com/questions/614750/understanding-n-left-frac2n-choose-n4n-right2-for-large-n ) — Winther, Sep 17 '19 at 17:01

score 1 · Answer 1 · answered Sep 17 '19 at 17:10

1

Prove it by induction on $n$. When $n$ increases to $n+1$ the right-hand side increases by a factor of ${4n\over n+1}$ What about the left-hand side?

answered Sep 17 '19 at 17:10

saulspatz

53,131

Prove $\binom{2n}{n}\geq\frac{4^n}{2n}$

1 Answers1