Binomial Inequality in which Binomial coefficient is in square root. 
$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$
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

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
$n = 100$.
