Prove this inequality for every $n\ge{1}$
$$\binom{2n}{n}\ge \frac{2^{2n}}{4\sqrt{n}+2} $$
I try to prove it with simplification this inequality but i don't find right path to get solution.
Asked
Active
Viewed 103 times
4
-
from where comes this problem? – Dr. Sonnhard Graubner Aug 12 '16 at 18:19
-
1HINT: Call $a_n$ the LHS, and $b_n$ the RHS. Try to prove that for all $n$ you have $$\frac{a_{n+1}}{a_n} \ge \frac{b_{n+1}}{b_n}$$ – Crostul Aug 12 '16 at 18:21
-
@Dr.SonnhardGraubner(Iranian mathematics Society)IMS compilation – Amir Aug 12 '16 at 18:28
-
hello Amir, glad to see you – Dr. Sonnhard Graubner Aug 12 '16 at 18:29
-
@Dr.SonnhardGraubner hello doctor ,nice to meet you to – Amir Aug 12 '16 at 18:32
-
1see http://math.stackexchange.com/questions/464721 for a proof using induction – user84413 Aug 12 '16 at 22:33
1 Answers
0
We begin with the product representations $${2n\choose n}{1\over 2^{2n}}={1\over 2n}\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)=\prod_{j=1}^n\left(1-{1\over2j}\right),\quad n\geq1.$$
From $$ \prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2}=\prod_{j=1}^{n-1}\left(1+{1\over j}+{1\over 4j^2}\right)\geq \prod_{j=1}^{n-1}\left(1+{1\over j}\right)=n,$$ we see that $$\left({2n\choose n}{1\over 2^{2n}} \right)^{2} = {1\over (2n)^2}\, \prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)^{\!\!2} \geq {1\over 4n^2}\, n ={1\over 4n},\quad n\geq1.$$ so by taking square roots, ${2n\choose n}{1\over 2^{2n}}\geq \displaystyle{1\over 2\sqrt{n}}\ge \frac{1}{4\sqrt{n}+2}.$
Amir
- 445