# Product and Square Root Proof

Let $a_1$ and $a_2$ be positive integers and let $m = a_1 a_2$.

Prove that at least one of $a_1$ or $a_2$ is at least $\sqrt m$.

Disclosure:

This is for a homework question, though the question is using the product of three numbers and the cubed root. I am looking more for some direction on solving this then an answer, I am hoping by better understanding how this would work with the product of two numbers I can work out doing this with three.

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Sketch made with paintroller: Assume otherwise, get contradiction. – anon Sep 29 '11 at 0:28
Consider the additive case: If $m = a_1 + a_2$, then at least one of $a_1$ or $a_2$ is at least $m/2$. – lhf Sep 29 '11 at 0:39

If neither is at least $\sqrt{m}$, then both are less than $\sqrt{m}$. If both $a_1$ and $a_2$ are less than $\sqrt{m}$, what does that say about $a_1 a_2$?

(In just the same way, but with the obvious simple changes, you could show that at least one of the two must be less than or equal to $\sqrt{m}$.)

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If both $a_1$ and $a_2$ are less than $\sqrt{m}$, then $a_1a_2\lt a_1\sqrt{m}\lt ?$

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$m$ can be thought of as product of two numbers $d\sqrt m$ and $\frac{\sqrt{m}}{d}$, where $d\geq0$. So in this way one number has to be greater than or equal to $\sqrt m$ and the other has to be less than or equal to it.

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