Non-existence of natural numbers such that $\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}$ Show that for any $n\in\mathbb{N}$ there does not exist natural numbers $x,y$ such that $$\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}.$$

Hint: Assume on the contrary that there is such $x, y$. Show $x+y \geq 2n+1$. Write $x+y=2n+1+k, k \geq 0$ and show $(2n+1-k)^2-1<4xy<(2n+1-k)^2$.
Hint 2: To show the last part, square the original inequality, shift some terms, and square again. Proving the lower bound will require slightly more work than the upper bound.

Edit: Since you appear to have so much difficulty with the lower bound, I guess it would be best to explain in more detail.
As mentioned in the second hint, square both sides and subtract $x+y$ to get
$$2n+1+2\sqrt{n}\sqrt{n+1}-(x+y)<2\sqrt{xy}<4n+2-(x+y)$$
Since $x+y=2n+1+k$, 
$$2\sqrt{n}\sqrt{n+1}-k<2\sqrt{xy}<2n+1-k$$
The upper bound trivially implies $k<2n+1$, so $k \leq 2n$, so $k<2\sqrt{n}\sqrt{n+1}$.
Thus the lower bound is positive, and we may square again.
$$(2\sqrt{n}\sqrt{n+1}-k)^2<4xy<(2n+1-k)^2$$
For the lower bound, 
\begin{align}
& (2\sqrt{n}\sqrt{n+1}-k)^2-((2n+1-k)^2-1)\\
& =(4n(n+1)+k^2-4k\sqrt{n}\sqrt{n+1})-(4n^2+4n+1+k^2-2k(2n+1)-1)\\
&=2k(2n+1-2\sqrt{n}\sqrt{n+1}) \\
&\geq 0
\end{align}
