Order theory between radicals

Question

Consider $p$ and $q$ $∈$ $N$ where $p>q$. What is the order between the numbers $\sqrt{2pq}$ and $\sqrt{p}+\sqrt{q}$?

Answer 1

It isn't clear what you are asking.

Might be worth remarking that, if $q≥2$ then: $$\sqrt {2pq}\;≥\;2\sqrt p\;>\,\sqrt p+\sqrt q$$ so the only counterexamples might occur with $q=1$.

Indeed, if $q=1$ then we can have $\sqrt {2p}<\sqrt p+1$. A little algebra shows that this works for $p\in \{1,2,3,4,5\}$. So, other than those $5$ counterexamples, $\sqrt {2pq}$ is always bigger than $\sqrt p+\sqrt q$.