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

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The discussion revolves around proving the inequality involving the sum of square roots of combinatorial coefficients from the expansion of (1+x)^n. The initial approach utilizes the RMS-AM inequality, leading to the expression that the sum is less than or equal to the square root of (2^n - 1) divided by n. Participants explore how to prove the relationship (2^n - 1)*n ≥ (2^n - 1)/n, confirming that n is a natural number greater than or equal to 1. The conversation concludes with one participant expressing gratitude for the assistance in reaching the solution. The discussion highlights the application of inequalities in combinatorial contexts.
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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!
 
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use
$$(2^n-1)*n\geq\frac{(2^n-1)}{n}$$
and then $$AM\geq GM$$
for $$2^n-1$$ and $$n$$
 
mathworker said:
use
$$(2^n-1)*n\geq\frac{(2^n-1)}{n}$$
and then $$AM\geq GM$$
for $$2^n-1$$ and $$n$$

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:
 
as $$C_0,C_1,C_2,...,C_n$$ are combinatorial coeffs of $$(1+x)^n$$. n is certainly an integer which is $$\geq 1 $$.
the rest is obvious...:)
 
mathworker said:
as $$C_0,C_1,C_2,...,C_n$$ are combinatorial coeffs of $$(1+x)^n$$. n is certainly an integer which is $$\geq 1 $$.
the rest is obvious...:)

Understood, thanks a lot mathworker! :)
 
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