<s>[INST] For $n \geq 2$, show that $\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$ Good day,
Can someone help me with giving hints for this problem:

Show that for $n \geq 2, n \in \mathbb{Z}$, $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$$

I tried $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} = \sum_{r = 1}^{n} \sqrt{r ^ 2\binom{n}{r}}$$
$$= \sqrt{n} \sum_{r = 1}^{n} \sqrt{r\binom{n - 1}{r - 1}}$$
So, we need to prove $$\sum_{r = 1}^{n} \sqrt{r\binom{n - 1}{r - 1}} < n \sqrt{2 ^ {n - 1}}$$
This reminds me of $$\sum_{r = 0}^{n} (r + 1) \binom{n}{r} = n 2 ^ {n - 1} + 2 ^ n = 2 ^ {n - 1}(n + 2)$$
but I can't see how to use it.
Also, the only complex math inequality I know is AM-GM, so it's quite possible that it is my theoretical knowledge that is lacking. So, if a well-known inequality is used, please also mention it.
Thanks
 [/INST] By Cauchy-Schwarz,
\begin{align*}
\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} &\le \left(\sum_{r = 1}^{n} r^2\right)^{1/2}\left(\sum_{r = 1}^{n} \binom{n}{r}\right)^{1/2} \\
&= \left((2^n-1)\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right)\right)^{1/2} \\
&< \left(2^{n-1}\cdot2\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right)\right)^{1/2}
\end{align*}
Then, verify $2\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right) < n^3$ for $n \ge 4$ by induction while manually checking the $n=2, n=3$ cases.
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