Let $\,u,v,x,y\,$ be real numbers. Prove that $ux+vy\leqslant\sqrt{u^2+v^2}\!\cdot\!\sqrt{x^2+y^2}$

Question

I am unsure how to go about this. This is a question for a mathematical proofs class at my University and the only proofs we have been exposed to so far are direct, contrapositive, and contradictory proofs. I tried squaring both sides but that seemed futile. I also tried proof by contradiction but it didn't seem to work. Any nudge in the right direction would be helpful.

Presumable one of the terms in the title should be $\sqrt{x^2+y^2}.$ It is certainly not true as written. Fixed, this is called the Cauchy-Schwarz inequality. — Thomas Andrews, Oct 16 '22 at 18:34
Squaring both sides and cancelling identical terms gives the equivalent inequality $2uxvy\le u^2y^2+v^2x^2$ or $0\le (uy-vx)^2$. — Kurt G., Oct 16 '22 at 19:50

score 2 · Answer 1 · answered Oct 16 '22 at 19:50

$ux\!+\!vy\leqslant\big|ux\!+\!vy\big|=\!\sqrt{\left(ux\!+\!vy\right)^2}\leqslant\!\sqrt{\left(ux\!+\!vy\right)^2\!+\!\left(uy\!-\!vx\right)^2}\!=$

$=\sqrt{u^2x^2+v^2y^2+u^2y^2+v^2x^2}=\sqrt{\left(u^2+v^2\right)\!\!\cdot\!\!\left(x^2+y^2\right)}=$

$=\sqrt{u^2+v^2}\!\cdot\!\sqrt{x^2+y^2}\,.$

Lelouch · Answer 2 · 2022-10-16T20:54:40.250

0

This is the the Cauchy-schwartz inequality in $\mathbb{R^2}$ with the euclidean product.

In this particular case, there is a geometric proof based on this figure (from Wikipédia) :

Otherwise you can look for any proof of CS inequality on the internet. Consider the polynomial of variable $t$ denoted by $P(t) = (x+tu)^2+(y+tv)^2 = (x^2+y^2)+2t(xu+yv)+t^2(u^2+v^2)$ and show that its discriminant is negative.

There is probably a more elementary approach for this particular case.

edited Oct 16 '22 at 20:54

answered Oct 16 '22 at 19:45

Lelouch

1,928

Actually, that determinant could be equal to $0$. – Angelo Oct 16 '22 at 20:01
This is included in the negative case (in my vocabulary at least). Anyway if t's 0 then there it's the equality case. – Lelouch Oct 16 '22 at 20:54
The word “negative” should mean “less than $0$”, whereas the word “nonpositive” should mean “less than or equal to $0$”. – Angelo Oct 16 '22 at 21:02
This is different in french – Lelouch Oct 16 '22 at 21:08
But in this site web we have to write words according to their English meaning, not French one. – Angelo Oct 16 '22 at 21:10
I am not aware of every subtilities of english conventions – Lelouch Oct 16 '22 at 21:14

Let $\,u,v,x,y\,$ be real numbers. Prove that $ux+vy\leqslant\sqrt{u^2+v^2}\!\cdot\!\sqrt{x^2+y^2}$

2 Answers2