# Help proving $\sum_{i=1}^n a_ib_i\leq \left(\sum_{i=1}^n a_i^2\cdot\sum_{i=1}^n b_i^2\right)^{1/2}$

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.

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

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.