number theory - Proofs from the BOOK: Bertrand's postulate: $binom{2m+1}{m}leq 2^{2m}$

I have a very hard proof from "Proofs from the BOOK". It's the section about Bertrand's postulate, page 8:

It's about the part, where the author says:

$$\binom{2m+1}{m}\leq 2^{2m}$$
because $\binom{2m+1}{m}=\binom{2m+1}{m+1}$ are the same in $\sum \limits_{k=0}^{2m+1} \binom{2m+1}{k}=2^{2m+1}$

I see, why they are the same, but I don't see the reason to say $\binom{2m+1}{m}\leq 2^{2m}$. Any help would be fine :)

Answer

The sum is essentially $a_1 + a_2 + ... +a_m + a_{m+1} + ... + a_{2m+1} = 2^{2m+1}$, (where $a_k$ is shorthand for $\binom{2m+1}{k}$) since everything is non-negative, we can say $a_m + a_{m+1} \leq a_1 + a_2 + ... +a_m + a_{m+1} + ... + a_{2m+1} = 2^{2m+1}$ and from the equality we know $a_m = a_{m+1}$. Putting it all together:

$a_m + a_{m+1} = 2a_m \leq 2^{2m+1}$ therefore $a_m \leq 2^{2m}$