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This question already has an answer here:

$\forall$:n∈${N}$

$\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$

I tried with no any success...

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marked as duplicate by Dietrich Burde, Juniven, kingW3, TastyRomeo, zhoraster Feb 18 '17 at 20:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $4^n = \sum \limits_{i=0}^{2n} \binom{2n}{i}$ and for all $i$, $\binom{2n}{i} \le \binom{2n}{n}$. If you manage to prove those two things, then you can derive the result you are looking for $\endgroup$ – charmd Feb 17 '17 at 11:26
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$$n=1 :\binom{2}{1}=2 \geq \dfrac{4}{3} \\n=k \to \binom{2k}{k}\geq \dfrac{4^k}{2k+1} \\ n=k+1 \to \binom{2(k+1)}{k+1}\geq \dfrac{4^{k+1}}{2(k+1)+1}$$ $$\binom{2(k+1)}{k+1}=\dfrac{(2k+2)!}{(k+1)!(k+1)!}=\dfrac{(2k+2)(2k+1)(2k)!}{(k+1)^2k!}=\\ \dfrac{(2k+2)(2k+1)}{(k+1)^2} \binom{2k}{k}\geq \dfrac{4.4^k}{2k+3}=4.\dfrac{2k+1}{2k+3}\dfrac{4^k}{2k+1}\\$$if we prove the $\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3}$ ,we proved the original relation $$\dfrac{(2k+2)(2k+1)}{(k+1)^2} \geq 4\dfrac{2k+1}{2k+3} \to \\ \dfrac{(2k+2)}{(k+1)^2} \geq 4\dfrac{1}{2k+3}\\ \dfrac{2}{(k+1)} \geq 4\dfrac{1}{2k+3}\\\\2(2k+3) \geq 4(k+1) \\4k+6 \geq 4k+4\\6\geq 4 \checkmark$$

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For $n=1$ we get $2\geq\frac{4}{3}$, which is true.

If $\binom{2n}{n}\geq\frac{4^n}{2n+1}$ we obtain: $$\binom{2n+2}{n+1}=\binom{2n}{n}\cdot\frac{(2n+2)(2n+1)}{(n+1)^2}>\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}$$ and it remains to prove that $$\frac{4^n}{2n+1}\cdot\frac{2(2n+1)}{n+1}>\frac{4^{n+1}}{2n+3}$$ or $$2(2n+3)>4(n+1),$$ which is obvious.

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