Induction on n?

any solution?

By Cauchy-Schwarz Inequality, we have $$(\frac{4n^{3/2}}{3} + \sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}})^2 \le (\frac{16}{9}n^3+2^2+2^4+...+2^{2n})(1+\frac{1}{1+0.5}+\frac{1}{2+0.5}+...+\frac{1}{n+0.5}).$$We proceed to show that $\frac{16}{9}n^3+2^2+2^4+...+2^{2n} < 2^{2n+2}$ and $1+\frac{1}{1+0.5}+\frac{1}{2+0.5}+...+\frac{1}{n+0.5} < n+1$. The latter is trivial as $$1+\frac{1}{1+0.5}+\frac{1}{2+0.5}+...+\frac{1}{n+0.5} < 1+1+1+...+1 = n+1.$$For the former, we just need to show $\frac{16}{9}n^3 < 2^{2n+1}$ since $2^2+2^4+...+2^{2n}<1+2+2^2+...+2^{2n}<2^{2n+1}$, which is always true since the left hand side is a polynomial function while the right hand side is an exponential function, thus the right hand side would increase faster than the left hand side. Hence, $$(\frac{4n^{3/2}}{3} + \sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}})^2 \le (\frac{16}{9}n^3+2^2+2^4+...+2^{2n})(1+\frac{1}{1+0.5}+\frac{1}{2+0.5}+...+\frac{1}{n+0.5}) < 2^{2n+2}(n+1).$$which when rearranged gives $$\sum_{k=1}^{n} \frac{2^k}{\sqrt{k+0.5}} \le 2^{n+1}\sqrt{n+1}-\frac{4n^{3/2}}{3},$$as desired.