Finding if sentence is true using induction: $\sqrt1 + \sqrt2 +\dots+\sqrt{n} \le \frac23(n+1)\sqrt{n+1}$ I have to prove
$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le  \frac{2}{3}*(n+1)\sqrt{n+1}$
by using math induction. 
First step is to prove that it works for n = 1 , which is true.
Next step is to prove it for n + 1. We can rewrite the formula using
$\sum_{i=1}^{n+1} \sqrt{i}= \sum_{i=1}^{n}\sqrt{i} + \sqrt{i+1}$
and we can substitute sum 
$\frac{2}{3}(n+1)\sqrt{n+1} +\sqrt{n+1} \le  \frac{2}{3}(n+2)\sqrt{n+2}$
we can transform the left side into 
$\sqrt{n+1}(\frac{2}{3}(n+1)+1)$
but how to I further transform the formula in order to find if the sentence is true?
Thanks for all help!

To finish your proof by induction we could use the fact the following inequalities are equivalent to each other
\begin{align*}
\frac23(n+1)^{3/2}+\sqrt{n+1} &\le \frac23(n+2)^{3/2}\\
\frac32\sqrt{n+1} &\le (n+2)^{3/2} - (n+1)^{3/2}\\
\frac32\sqrt{n+1} &\le (\sqrt{n+2} - \sqrt{n+1}) (n+2 + \sqrt{(n+2)(n+1)} + n+1)\\
\frac32\sqrt{n+1}(\sqrt{n+2} + \sqrt{n+1}) &\le n+2 + \sqrt{(n+2)(n+1)} + n+1\\
\frac32(n+1 + \sqrt{(n+2)(n+1)}) &\le 2n+3 + \sqrt{(n+2)(n+1)}\\
\frac12\sqrt{(n+2)(n+1)} &\le \frac{n+3}2
\end{align*}
I have:
Used $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=\sqrt{n+2}$ $b=\sqrt{n+1}$
Multiplied both sides by $\sqrt{n+2}+\sqrt{n+1}$ to use that $(\sqrt{n+2} - \sqrt{n+1})(\sqrt{n+2} + \sqrt{n+1}) = (n+2)-(n+1)=1$.

If you already know how to integrate, you can simply use that 
$$\newcommand{\dx}{\; \mathrm{d}x}\sqrt k \le \int_{k}^{k+1} \sqrt x \dx.$$
Simply by adding these inequalities together for $k=0$ to $n$ you get
$$ \sum_{k=0}^n \sqrt n \le \int_0^{n+1} \sqrt x \dx = \frac{(n+1)^{3/2}}{3/2} = \frac23 (n+1)\sqrt{n+1}.$$
This can be also visualized by noticing that one side is are of rectangle under the curve $y=\sqrt x$ and the other is the whole area under the curve.
I will add the following - somewhat similar - picture which might help (it was taken from this post):

