Prove using the smallest counterexample technique that: $\binom {2n}n\leqslant4^n.$

Question

Actually what i know is that i must assume that this statement is false and then try to come up with non sense statement.

Prove by the smallest counterexample technique the statement $$\binom {2n}n\leqslant4^n.$$

Answer 1

Assume that $$\binom{2n}{n}>4^n$$ for some $n\in \mathbb{N}.$ Then

$$\binom{2n}{n}=\frac{2(2n-1)}{n}\binom{2(n-1)}{n-1}>4^{n-1}\implies \binom{2(n-1)}{n-1}>\frac{n4^n}{2(2n-1)}>4^{n-1}.$$

That is, if the inequality fails to hold for some $n\in \mathbb{N}$ then it fails to hold for any smaller natural. In particular, we would have

$$2=\binom{2}{1}>4^1=4,$$

which gives the desired contradiction.