⌊√n + √(n+1)⌋ = ⌊√(4n+2)⌋

Idea. The two numbers are very close; an integer between them would be a square near $4n+2$, and $4n+2$ is never a square.

Upper comparison. $(\sqrt n+\sqrt{n+1})^2=2n+1+2\sqrt{n(n+1)}$, and $n(n+1)<\left(n+\frac12\right)^2$, so $2\sqrt{n(n+1)}<2n+1$ and
$$(\sqrt n+\sqrt{n+1})^2<4n+2.$$

Lower comparison. $n(n+1)>n^2$ so $2\sqrt{n(n+1)}>2n$ and
$$(\sqrt n+\sqrt{n+1})^2>4n+1.$$

No integer in between. Suppose an integer $k$ satisfies $\sqrt n+\sqrt{n+1}<k\le\sqrt{4n+2}$. Squaring, $4n+1<k^2\le4n+2$, so $k^2=4n+2$. But squares are $\equiv0,1\pmod4$ while $4n+2\equiv2$. Contradiction.

Conclusion. The numbers $\sqrt n+\sqrt{n+1}\le\sqrt{4n+2}$ have no integer in $(\sqrt n+\sqrt{n+1},\sqrt{4n+2}]$, so their floors coincide.

Check. $n=1$: $1+1.414=2.414$, $\sqrt6=2.449$, both floors $2$.

Lesson. To compare floors of two close irrational expressions, show that no integer (or the square of an integer) can separate them, often with a congruence argument.