# Let $p$=prime and $\sqrt{x}+\sqrt{y}<\sqrt{2p}$

Let $p$ be a fixed odd prime. Let $x,y\in \mathbb{Z}_+$ such that $\sqrt{x}+\sqrt{y}<\sqrt{2p}$. Prove that $$\sqrt{x}+\sqrt{y}\le \sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}.$$

Any ideas at all? This seems extremely difficult to do using elementary methods.

Note: It is from the 2015 Moldova TST (IMO selection test). The original problem was: Let $p$ be a fixed odd prime and $x,y\in \mathbb{Z}_+$. Find the minimum positive value of $\sqrt{2p}-\sqrt{x}-\sqrt{y}$.

EDIT! : This is apparently an old IMO Shortlist problem which Moldova recycled for their TST. I note that my reformulation of it was correct; see here for a solution: See here.

• The first thing I'd try is squaring everything. If $\sqrt x + \sqrt y < \sqrt{2p}$, is it necessarily also true $x + y < 2p$? This is just an idea, I haven't thought it all the way through. – John-Luke Mar 31 '15 at 21:07
• You may want to write your answer in the comments so that this question is marked as answered. Relevant meta: meta.math.stackexchange.com/questions/1559/… – Element118 Nov 14 '15 at 8:50
• Here should be proved a more general claim. – Alex Ravsky May 29 '17 at 1:12

Let's suppose that the problem is not true and there exists $x,y$ such that $$\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}<\sqrt{x}+\sqrt{y}<\sqrt{2p}\qquad (1)$$ First, note that if $x+y\le p$ then $$\sqrt{x}+\sqrt{y}\le\sqrt{x}+\sqrt{p-x}\le\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}$$ Since $\sqrt{x}+\sqrt{p-x}$ is concave function and maximal at $x=p/2$. Squaring both sides of $(1)$ gives $$p+\sqrt{p^2-1}<x+y+2\sqrt{xy}<2p$$ Now, because we showed $x+y>p$, let $\epsilon=p-\sqrt{p^2-1}$ then it is enough to prove that there is no integer $n, m$ with $m<p$ and $$m-\epsilon<2\sqrt{n}<m\qquad (2)$$ Now, suppose that $m$ is even and there exists $k$ such that $2k=m$. Then we get $k<p/2$, also $n<k^2$, which is $n\le k^2-1$ and $$2k-\epsilon <2 \sqrt{n}\le 2\sqrt{k^2-1}$$ $$2(k-\sqrt{k^2-1})<p-\sqrt{p^2-1}$$ However, since $f(x)=\sqrt{x}-\sqrt{x-1}$ is strictly decreasing function for positive $x$, therefore \begin{align}2(k-\sqrt{k^2-1})& > 2\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}-1}\right)\\& =p-\sqrt{p^2-4}\\& > p-\sqrt{p^2-1}\end{align} And this is contradiction.
Now we suppose that $m$ is odd and there exists $k$ such that $m=2k+1$. Also we get $4n < 4k^2+4k+1$, so $n \le k^2+k$. Also, from $2k+1-\epsilon<2\sqrt{n}\le2\sqrt{k^2+k}$,$$2k+1-2\sqrt{k^2+k}<p-\sqrt{p^2-1}$$ However, this cannot be true, because \begin{align}2k+1-2\sqrt{k^2+k}& = 2k+1-\sqrt{(2k+1)^2-1}\\& =m-\sqrt{m^2-1}\\& > p-\sqrt{p^2-1}\end{align}
Therefore, this is contradiction and there exists no integers $m,n$ with $(2)$, and it follows that there exists no integers $x,y$ satisfying $(1)$. Proved!