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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.

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    $\begingroup$ You are not allowed to use Cauchy-Schwarz inequality ? Because this is basically it $\endgroup$ – Tuvasbien Apr 20 at 1:19
  • $\begingroup$ If I used it wouldn't that be circular reasoning? $\endgroup$ – drfrankie Apr 20 at 1:21
  • $\begingroup$ What you want to prove is the Cauchy-Schwarz inequality, so you want a proof of this inequality if I understand ? $\endgroup$ – Tuvasbien Apr 20 at 1:22
  • $\begingroup$ I've seen how the CS inequality is proven. Is that all I need to clear case 2? $\endgroup$ – drfrankie Apr 20 at 1:27
  • $\begingroup$ No need to have different cases, Cauchy-Schwarz inequality is exactly what you want to prove. What is the Cauchy-Schwarz inequality for you ? $\endgroup$ – Tuvasbien Apr 20 at 1:29
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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.

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