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.
 A: 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!
