# 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}

• But how to show that $(2n+1-k)^2 -1 <4xy<(2n+1-k)^2$?
– user110661
Jun 4, 2014 at 15:52
• @FisiaiLusia Have you shown $x+y \geq 2n+1$? What have you tried in relation to my hint? Jun 4, 2014 at 15:54
• Yes, if $x+y < 2n+1$ then $$\sqrt{x} +\sqrt{y} <\sqrt{2n+1 +2\sqrt{xy}} \leq \sqrt{n} +\sqrt{n+1} .$$
– user110661
Jun 4, 2014 at 16:04
• Oh I see, if $x+y =2n+1+k$ and $\sqrt{x} +\sqrt{y} <\sqrt{4n+2}$ then $$2\sqrt{xy} <4n+2 -(x+y) =2n+1-k$$ hence $$4xy< (2n+1-k)^2$$
– user110661
Jun 4, 2014 at 16:10
• But, how to show that $4xy>(2n+1-k)^2 -1$?
– user110661
Jun 4, 2014 at 16:15