Non-existence of natural numbers such that $\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}$

Question

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

Answer 1

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.

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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}