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Proving an inequality

Pranav

Well-known member
Nov 4, 2013
428
Problem:
Prove:
$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+......+\sqrt{C_n} \leq 2^{n-1}+\frac{n-1}{2}$$

where $C_0,C_1,C_2,....,C_n$ are combinatorial coefficients in the expansion of $(1+x)^n$, $n \in \mathbb{N}$.

Attempt:

I thought of using the RMS-AM inequality and got:

$$\sqrt{C_1}+\sqrt{C_2}+\sqrt{C_3}+.....+\sqrt{C_n} \leq \sqrt{\frac{2^n-1}{n}}$$

Where I have used $C_1+C_2+C_3+.....+C_n=2^n-1$.

But I don't see how to proceed from here.

Any help is appreciated. Thanks!
 
Last edited:

mathworker

Active member
May 31, 2013
118
use
\(\displaystyle (2^n-1)*n\geq\frac{(2^n-1)}{n}\)
and then \(\displaystyle AM\geq GM\)
for \(\displaystyle 2^n-1\) and \(\displaystyle n\)
 

Pranav

Well-known member
Nov 4, 2013
428
use
\(\displaystyle (2^n-1)*n\geq\frac{(2^n-1)}{n}\)
and then \(\displaystyle AM\geq GM\)
for \(\displaystyle 2^n-1\) and \(\displaystyle n\)
Thanks mathworker!

I was able to reach the answer but how do you prove $(2^n-1)*n\geq\frac{(2^n-1)}{n}$? :confused:
 

mathworker

Active member
May 31, 2013
118
as \(\displaystyle C_0,C_1,C_2,....,C_n\) are combinatorial coeffs of \(\displaystyle (1+x)^n\). n is certainly an integer which is \(\displaystyle \geq 1 \).
the rest is obvious......:)
 

Pranav

Well-known member
Nov 4, 2013
428
as \(\displaystyle C_0,C_1,C_2,....,C_n\) are combinatorial coeffs of \(\displaystyle (1+x)^n\). n is certainly an integer which is \(\displaystyle \geq 1 \).
the rest is obvious......:)
Understood, thanks a lot mathworker! :)