Let $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$ be real numbers. Prove that $$\sum_{i=1}^n a_ib_i\leq \left(\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n b_i^2\right)^{1/2}$$
First, I recognized that $A \leq B^{1/2}$ does not imply $A^2\leq B$, where $A=\sum_{i=1}^n a_ib_i$ and $B=\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n b_i^2$. Instead it must be separated into two cases, either $A\leq|B|$ or $A<-|B|$. I proved case 1 using induction. Now, without using the Cauchy-Schwarz inequality I need to show that $A^2>B$ is impossible which would take care of case 2.
Am I even approaching this right? I am currently in an introductory proofs class for context.
My initial suggestion would be to use the usual proof of Cauchy-Schwarz, which is really neat (and it applies in way more generality than this).
If you want to fight your proof "by hand", the first thing to notice is that, since $$ \sum_i a_ib_i\leq\sum_i|a_i|\,|b_i| $$ and the right-hand-side does not see the signs, you can assume without loss of generality that $a_i\geq0$, $b_i\geq0$ for all $i$. If you have prove this, as you say, you are done.