This was my solution on the test

I tried ceiling function, i tried AM-GM, I tried for constructing m. But i tried all of them separately, and you combined them all. Great solution

I did the same

But i wrote m = root ab +1 is root ab is an integer. I could give better sol

Purple_Planet wrote: Given positive real numbers $a, b, c, d$ such that $cd=1$. Prove that there exists at least one positive integer $m$ such that $$ab\le m^2\le (a+c) (b+d). $$ Trivial by Cauchy-Schwarz!

We have $(bc - ad)^2 \geq 0$ by trivial inequality. $$b^2 + a^2d^2 - 2abcd \geq 0$$Adding $4abcd$ to both sides, we have $b^2c^2 + a^2d^2 + 4abcd \geq 4abcd$, that is, $(bc + ad)^2 \geq 4abcd$. Taking the square root of both of sides, we get: $$bc + ad \geq 2\sqrt{ab}$$We can add $1 + ab$ to both sides to get $bc + ad + ab + 1 \geq 2\sqrt{ab} + ab + 1$, or $bc + ad + ab + cd \geq 2\sqrt{ab} + ab + 1$. Note that $bc + ad + ab + cd = (a + c)(b + d)$ and that $2\sqrt{ab} + ab + 1 = (\sqrt{ab} + 1)^2$. We can replace this to get that $(a + c)(b + d) \geq (\sqrt{ab} + 1)^2$. Taking the square root of both sides, we have $\sqrt{(a + c)(b + d)} \geq \sqrt{ab} + 1$, or $\sqrt{(a + c)(b + d)} - \sqrt{ab} \geq 1$. Let $p = \sqrt{(a + c)(b + d)}$ and let $q = \sqrt{ab}$. As $p - q \geq 1$, we have $q^2 \leq (q + 1)^2 \leq p^2$. We are done.

$(a+c)(b+d) = ab+ad+bc+1 \ge ab+2\sqrt(ab)+1=(\sqrt(ab)+1)^2 \ge ( [ \sqrt(ab) ]+1)^2 \ge ab$ Thus $m= [ \sqrt(ab) ]+1$ satisfies.