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

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

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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


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 $n = 100$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.