Talk:Triangle Inequality/Complex Numbers/Proof 2

The proof is sound, I think. If |a1| > |b1|, then we would continue with (a1^2+b2^2) and after taking the left hand side of the inequality to the right, we would factor out a1 and b2. Similarly for the other two cases.

So we can either assume w.l.o.g. (thanks for reformatting, your edit makes it much better) or we can treat each case.

I did not notice the material was quoted. Maybe that's why my professor was so dismissive of the Schaum series. That said, I only have a bachelor's (I graduated in 1989, but my grades were not stellar) and came back to math as a hobby 10 years ago. I was browsing G.H. Hardy's textbook last week and came upon that problem on page 89 of the Dover edition. I figured it would be simple, spent two hours working on it and I came up with the same proof Spiegel did!!! Anyway, I wrote a fair copy longhand, took a picture of my work, posted it on Facebook, where a friend pointed the mistake (-6 < 5, but 36 > 25). D'oh! I felt like Homer Simpson (I'm not as smart as I'd like to be) so I spent another two hours fixing it.

I found Spiegel's proof on Amazon, but I still think my friend's comment is correct and that Spiegel's proof is wrong. https://www.amazon.ca/Schaums-Outline-Complex-Variables-2ed/dp/0071615695/ref=sr_1_9?dchild=1&keywords=Murray+Spiegel&qid=1594986935&sr=8-9

Vincent - Vfp15 (talk) 12:31, 17 July 2020 (UTC)

Spiegel's proof is sound if we can demonstrate that $\size {a_1 b_1 + a_2 b_2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$.

This appears to be what you yourself are working towards in your own analysis -- but as we note, there is a missing case which needs treatment.

I wonder whether it's worth trying to isolate that specific step as a lemma, and then invoke it during the course of continuing the proof as per Spiegel from that point?

Incidentally, my own formal mathematics education suggested one or two of the Schaum manuals for practice problems and use as general workbooks, but we were to beware the errors that were all too frequent. --prime mover (talk) 13:15, 17 July 2020 (UTC)

It occurs to me we can use Cauchy's Inequality, which gives $\paren {a_1 b_1 + a_2 b_2}^2 \le \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2}$ directly.

So the original proof has been restored. --prime mover (talk) 13:51, 17 July 2020 (UTC)