Proving an inequality

Pranav

Well-known member
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

Well-known member
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
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 $(2^n-1)*n\geq\frac{(2^n-1)}{n}$?

mathworker

Well-known member
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
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!