How can I prove this inequality for $n\geq 2$?

Question

How can I prove this inequality for all natural numbers $n\geq 2$?

${2n\choose n}>\frac{4^n}{n+1}$

I've tried induction but that was a dead end.

Answer 1

For $n=2$, $${4\choose 2}=6> \frac{16}{3}$$

Assume $${{2k}\choose{k}}>\frac{4^k}{k+1}$$

Then \begin{align} \binom{2k+2}{k+1}&=\frac{(2k+2)(2k+1)}{(k+1)^2}\binom{2k}{k}\\ &>\frac{(2k+2)(2k+1)}{(k+1)^2}\left(\frac{4^k}{k+1}\right)\\ &>\frac{4(k+1)}{k+2}\left(\frac{4^k}{k+1}\right)\\ &=\frac{4^{k+1}}{k+2}. \end{align}

In the middle there we used $$\frac{2(2k+1)}{k+1}>\frac{4(k+1)}{k+2}$$ which is equivalent to $k(k+1)(k+2)>0$.

Answer 2

Induction actually does work. Let $a_n$ denote the LHS and $b_n$ denote the RHS at the $n^{\text{th}}$ step. Then $$ \frac{a_n}{a_{n-1}} = \frac{2n(2n-1)}{n^2} = \frac{4n-2}n, \quad \frac{b_n}{b_{n-1}} = \frac{4n}{n+1}. $$ To figure out which factor is bigger, we consider their ratio : $$ \frac{\frac{4n-2}{n} }{ \frac{4n}{n+1}} = \frac{(4n-2)(n+1)}{4n^2} = \frac{4n^2+2n-2}{4n^2} \ge 1 $$ as long as $2n-2 \ge 0$, i.e. as long as $n \ge 1$. So you can start induction at $n=2$ where $\binom{2n}n = \binom 42 = 6 > \frac{16}3 = \frac{4^2}{2+1}$.

Hope that helps,

Answer 3

Recall that for any real numbers $x_0,x_1,\ldots,x_n$ the following inequality holds: $$ x_0^2+x_1^2+\ldots+x_n^2\ge\frac{1}{n+1}(x_0+x_1+\ldots+x_n)^2. $$ This is a consequence of the Cauchy-Schwarz inequality, for example. Note that the equality holds if and only if all $x_k$'s are equal.

Now apply this inequality for $x_k=\binom{n}{k}$, where $k\in\{0,1,\ldots,n\}$ and use the following well-known combinatorial identities: $$ \binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n}=2^n,~\text{and}\\ \binom{n}{0}^2+\binom{n}{1}^2+\ldots+\binom{n}{n}^2=\binom{2n}{n}. $$ We obtain $$ \binom{2n}{n}\ge\frac{1}{n+1}(2^n)^2=\frac{4^n}{n+1}. $$ Finally, observe that for $n\ge 2$ the equality cannot hold since $x_0=\binom{n}{0}=1$ while $x_1=\binom{n}{1}=n>1$. Thus, the inequality is strict for $n\ge 2$.

Remark. I think I saw this problem elsewhere some time ago. I posted the solution that a friend of mine came up with back then.

Answer 4

Although it might look astounding (at least to me), the following equation holds: $$\begin{align} \binom{2n}{n}&=\frac{(2n)!}{n!\cdot n!}=\prod_{k=1}^n\frac{2k(2k-1)}{k\cdot k} =2^{2n}\prod_{k=1}^n\frac{2k(2k-1)}{2k\cdot2k} =2^{2n}\prod_{k=1}^n\left(1-\frac{1}{2k}\right) \end{align}$$ Now, for $k\ge1$, we have $2k\ge k+1$ and so $1-\frac{1}{2k}\ge1-\frac{1}{k+1}$ where we only have equality if $k=1$. Therefore, if $n\ge2$: $$\begin{align} \binom{2n}{n}>2^{2n}\prod_{k=1}^n\left(1-\frac{1}{k+1}\right)&=2^{2n}\prod_{k=1}^n\frac{k}{k+1}=\frac{2^{2n}}{n+1} \end{align}$$

Answer 5

Consider $$p(x)=(x-1)(x-3)...(x-2n+1)$$ which is concave up for $x>2n-1.$ So, $$\frac{p(2n)+p(2n+2)}2>p(2n+1)$$ $$(n+1)(1.3...(2n-1))>2.4...(2n)$$