# Inequality, prove for all natural numbers $n \geq 2$, possibly induction

Prove that for every natural number $n \geq2$ there is inequality $$1\cdot\sqrt{\binom{n}{1}}+2\cdot\sqrt{\binom{n}{2}}+3\cdot\sqrt{\binom{n}{3}}+...+n\cdot\sqrt{\binom{n}{n}}<\sqrt{2^{n-1}\cdot n^3}$$ I'm interested in various ways this problem can be solved. I guess it should be solvable by induction because there was in the content of the problem prove for all natural $n>2$ . Hints on various solutions are welcome. Also I tried normal induction and it's too difficult to solve (unless there exists a way to simplify it somehow). And I tried to use QM-AM and then induction on it and that's difficult too.
EDIT: I could have made mistake with this QM-AM induction.

Using Cauchy-Schwartz we have \begin{equation} \bigg(1\cdot\sqrt{\binom{n}{1}}+2\cdot\sqrt{\binom{n}{2}}+3\cdot\sqrt{\binom{n}{3}}+...+n\cdot\sqrt{\binom{n}{n}}\bigg)^2\leq (1^2+2^2+...+n^2)2^n. \end{equation} Since \begin{equation} 1^2+2^2+\dots+n^2 = \frac{1}{6}n(n+1)(2n+1)<n^3/2. \end{equation} for $n>3$. I think we are done.
QM-AM $$\frac{1\cdot\sqrt{\binom{n}{1}}+2\cdot\sqrt{\binom{n}{2}}+3\cdot\sqrt{\binom{n}{3}}+\dots+n\cdot\sqrt{\binom{n}{n}}}{n}\leq\sqrt{\frac{1^2\cdot\sqrt{\binom{n}{1}}+2^2\cdot\sqrt{\binom{n}{2}}+3^2\cdot\sqrt{\binom{n}{3}}+\dots+n^2\cdot\sqrt{\binom{n}{n}}}{n}}$$ Now this is what we're trying to prove. $$n\cdot\bigg(1^2\cdot\sqrt{\binom{n}{1}}+2^2\cdot\sqrt{\binom{n}{2}}+3^2\cdot\sqrt{\binom{n}{3}}+\dots+n^2\cdot\sqrt{\binom{n}{n}}\bigg)<2^{n-1}\cdot n^3$$ Next is the induction step $$2^{n-1}\cdot n^2+(n+1)^2 < 2^n \cdot (n+1)^2$$ We can rewrite it then as: $$0<n^2+\bigg(1+\frac{2^{n-1}}{2^{n-1}-1}\bigg)(2n+1)$$ Which is true because we had to prove it for every $n\geq2$ $$\frac{2^{n-1}}{2^{n-1}-1}>1$$ Can someone verify it?