Can the RMS-AM inequality prove the combinatorial coefficient inequality?

[Saitama]

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!

 

[mathworker]

use
\(\displaystyle (2^n-1)*n\geq\frac{(2^n-1)}{n}\)
and then \(\displaystyle AM\geq GM\)

Spoiler

for \(\displaystyle 2^n-1\) and \(\displaystyle n\)

 

[Saitama]

use
\(\displaystyle (2^n-1)*n\geq\frac{(2^n-1)}{n}\)
and then \(\displaystyle AM\geq GM\)

Spoiler

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]

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...:)

 

[Saitama]

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! :)