Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$

We may consider that $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence for any $n\in\mathbb{N}^+$ we have $$ f(n-1/2)-f(n+1/2) \geq -f'(n) $$ or $$ \frac{1}{\sqrt{n-1/2}}-\frac{1}{\sqrt{n+1/2}} \geq \frac{1}{2n\sqrt{n}} $$ such that by creative telescoping $$ \sum_{n=1}^{N}\frac{1}{n\sqrt{n}} \leq 2\left(\frac{1}{\sqrt{1-1/2}}-\frac{1}{\sqrt{N+1/2}}\right) < 2\sqrt{2}.$$ The inequality is pretty sharp since $\zeta\left(\frac{3}{2}\right)=2.612375\ldots$
The same approach proves $\zeta\left(\frac{3}{2}\right)\leq 1+2\sqrt{\frac{2}{3}}<2.633$.

Your sum can be written as $\sum_{k=1}^n\frac{1}{k^{3/2}}=1+2^{-3/2}+\sum_{k=3}^n k^{-3/2}$. Since $f(x)=x^{-3/2}$ is decreasing, we can bound the sum from above by an integral, that is,

$$ \sum_{k=3}^n\frac{1}{k^{3/2}}\leq\int_2^n x^{-3/2}\,dx. $$ Note specifically the lower bound of the integral.

Evaluating the integral on the right yields $$ \int_2^n x^{-3/2}\,dx=-2x^{-1/2}\bigg|_{x=2}^{x=n}=\frac{2}{\sqrt2}-\frac{2}{\sqrt n}. $$ Thus, we have $$ \sum_{k=1}^n\leq1+2^{-3/2}+\frac{2}{\sqrt2}-\frac2{\sqrt n}\leq\frac{2^{3/2}+1+4}{2^{3/2}}. $$ Clearly $2^{3/2}=2\sqrt2=\sqrt8\leq\sqrt9=3$, so that we have $$ \frac{2^{3/2}+1+4}{2^{3/2}}\leq\frac{8}{2^{3/2}}=2\sqrt2. $$ Putting it all together, then, we have $$ \sum_{k=1}^nk^{-3/2}\leq2\sqrt2. $$

I think your way gets something wrong.

By your work:

For all $n\geq2$ we obtain: $$\frac{1}{n\sqrt{n}}<\left(\frac{1}{\sqrt{n-1}}-\frac{1} {\sqrt{n}}\right)\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}=$$ $$=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\sqrt{\frac{n}{n-1}}\right)\leq\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)(1+\sqrt2).$$ Thus, $$\sum_{k=1}^n\frac{1}{k\sqrt{k}}=1+\sum_{k=2}^n\frac{1}{k\sqrt{k}}\leq1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right).$$ Id est, it's enough to prove that $$1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right)<2\sqrt2,$$ which is wrong for $n\rightarrow+\infty$.