<s>[INST] For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$
 [/INST] Case 1: $n^2\le x\le(n+1)^2-2$ (for $n\ge 3$), then
$$ n\le \sqrt{x}\le\sqrt{(n+1)^2-2}<n+1, n<\sqrt{n^2+1}\le \sqrt{x+1}<\sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<\sqrt{4n^2+2}\le \sqrt{4x+2}<\sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<\sqrt{x}+\sqrt{x+1}<2n+2, \sqrt{4x+2}=2n $$
and
$$ 2n+1\le[\sqrt{x}+\sqrt{x+1}]\le2n+2, [\sqrt{4x+2}]=2n+1. $$
Now we show
$$ \sqrt{x}+\sqrt{x+1}<2n+1. $$
Note $\sqrt{x}+\sqrt{x+1}$ is increasing and hence
$$\sqrt{x}+\sqrt{x+1}\le\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}.$$
Note
\begin{eqnarray}
&&\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}-2n-1\\
&=&\frac{(n+1)^2-2-n^2}{\sqrt{(n+1)^2-2}+n^2}+\frac{(n+1)^2-1-n^2}{\sqrt{(n+1)^2-1}+n^2}-1\\
&<&\frac{4n-1}{\sqrt{(n+1)^2-2}+n^2}-1\\
&=&\frac{4n-1}{\sqrt{n^2+2n-1}+n^2}-1\\
&<&\frac{4n-1}{n+n^2}-1\\
&=&-\frac{n^2-3n+1}{n+n^2}\\
&<0
\end{eqnarray}
for $n\ge 3$.
Thus 
$$2n<\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}<2n+1$$
or 
$$[\sqrt{x}+\sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
\begin{eqnarray}
&&\sqrt{x}+\sqrt{x+1}\\
&=&\sqrt{(n+1)^2-1}+n+1\\
&<&2(n+1)
\end{eqnarray}
and
\begin{eqnarray}
&&\sqrt{x}+\sqrt{x+1}\\
&=&\sqrt{(n+1)^2-1}+n+1\\
&>&2n+1
\end{eqnarray}
and hence
$$ [\sqrt{x}+\sqrt{x+1}]=2(n+1). $$
On the other hand
$$ \sqrt{4x+2}=\sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ \sqrt{4x+2}=\sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [\sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
</s>