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$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$

where $\displaystyle C_{r}=\binom{n}{r}$

Try: Using Cauchy Schwarz Inequaity

$$\bigg(1^2+2^2+\cdots \cdots +100^2\bigg)\bigg(C_{1}+C_{2}+\cdots \cdots +C_{100}\bigg)\geq \bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)^2$$

$$\bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)\leq \bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}$$

i am not understand how can i prove my original inequality,

could some help me , Thanks

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    $\begingroup$ What does $C_n$ mean? $\endgroup$ – Hw Chu May 11 '18 at 18:06
  • $\begingroup$ I think, $C_n=\binom{100}{n}$. $\endgroup$ – DiegoMath May 11 '18 at 18:08
  • $\begingroup$ I think you meant $(C_1+\cdots+C_{100})$ in your application of Cauchy Schwartz. $\endgroup$ – Alex R. May 11 '18 at 18:20
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    $\begingroup$ The inequality seemed to be very loose. Try comparing terms of the form $r\sqrt{C_r}$, we can estimate that the maximal one happens at $r_m = 51$ or $52$. So the left hand side is less than $100\times r_m\sqrt{C_{r_m}}$. But $\sqrt{C_{r_m}} < 2^{50}$, so the entire stuff is less than $5200\times 2^{50}$. $\endgroup$ – Hw Chu May 11 '18 at 18:21
  • $\begingroup$ Thanjs Hw Chu i did not understand , please explain me in detail. $\endgroup$ – DXT May 11 '18 at 18:27
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Hint.

$$ 2\bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}\le \frac{100\cdot 101\cdot 201}{6} + (2^{100}-1) $$

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By C-S, as you have done,

$\begin{array}\\ \left(\sum_{k=1}^n k\sqrt{\binom{n}{k}}\right)^2 &\le \sum_{k=1}^n k^2\sum_{k=1}^n\binom{n}{k}\\ &=\dfrac{n(n+1)(2n+1)}{6}(2^n-1)\\ &\lt\dfrac{2(n+1)^3}{6}2^n\\ &=\dfrac{(n+1)^3}{3}2^n\\ \text{so}\\ \sum_{k=1}^n k\sqrt{\binom{n}{k}} &\lt\dfrac{(n+1)^{3/2}}{\sqrt{3}}2^{n/2}\\ \end{array} $

and this is less than $2^{n-1}$ for $n \ge 12$ according to Wolfy.

So it is much less for your case of $n = 100$.

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