Inequality with sum of Binomial coefficients: $\sum\limits^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}$

Question

Prove that for every positive integer $n \ge 2$ $$\sum^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}$$ I tried it by induction but I didn't know how to end it.

Answer 1

As mentioned in the comments we can get bounds using CS inequality. Here is a tighter bound using CS and appropriately chosen sequences. First note that: $$\sum_{k=0}^n\binom{n}{k}x^k = (1+x)^n \implies \sum_{k=0}^nk \binom{n}{k}x^k = x\frac{d}{dx}(1+x)^n = n x(1+x)^{n-1}$$

$$\implies \sum_{k=1}^nk \binom{n}{k} = n 2^{n-1} $$

Now by CS inequality: $$\left(\sum_{k=1}^nk \right) \cdot \left(\sum_{k=1}^nk \binom{n}{k}\right) \ge \left(\sum_{k=1}^n k \sqrt{\binom{n}k} \right)^2$$ $$\implies \frac{n(n+1)}2\cdot n 2^{n-1} \ge \left(\sum_{k=1}^n k \sqrt{\binom{n}k} \right)^2 \implies \sum_{k=1}^n k \sqrt{\binom{n}k} \le n 2^{n/2-1} \sqrt{n+1}$$

P.S. That this bound is tighter than your RHS is easily shown: $$n 2^{n/2-1} \sqrt{n+1} \le \sqrt{2^{n-1}n^3} \iff n+1 \le 2n \iff n \ge 1$$