Quadratic equations and inequalities $\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$ and $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$ For every positive integer $n$, prove that 
               $$\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$$
Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$, where $[x]$ denotes the greatest integer not exceeding $x$. 
This question was posed to me in class by my teacher. Since we are discussing quadratic equations. I can only imagine that this question is related to that topic.
Actually, by squaring the terms on both sides of the inequality, the first part of the question is solved easily. It is the second half that is causing me trouble.
Clearly we have to show, that if $x <\sqrt{4n+1} < x+1$ where $x$ is a natural number, then $x <\sqrt{n} + \sqrt{n+1} < x+1$, but how?
I am in high school, so please use techniques appropriate for my level.

Observe that $$\lfloor\sqrt{4n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$$
unless $4n+2$ is perfect square
But any square $\equiv0,1\pmod4$  
$$\implies\lfloor\sqrt{4n+1}\rfloor=\lfloor\sqrt n+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$$
