Hello, can you help me with this inequality?

Question

for $x,y,u,v\in R$ $$|ux-yv|\leq\sqrt{(x^{2}+y^{2})\times(u^{2}+v^{2})}$$. Thank you

Answer 1

If $\vec x=(y,x,0)$ and $\vec u=(u,v,0)$, and $\theta$ is the angle between them, then $|ux-yv|=|\vec x\times \vec u|=\|\vec x \|\|\vec u\||\sin \theta|\le \|\vec x \|\|\vec u\|=\sqrt{(x^{2}+y^{2})\cdot (u^{2}+v^{2})}$

Answer 2

$|ux - vy| = \sqrt {(ux - vy)^2}$ by definition.

$\sqrt {(ux - vy)^2} \le \sqrt{(x^2 + y^2)(u^2+v^2)}$

Squaring both sides

$(ux - vy)^2 \le (x^2 + y^2)(u^2+v^2)$

Expanding the binomials

$u^2x^2 - 2uvxy + v^2y^2 \le u^2x^2 + u^2y^2 + x^2v^2 + y^2v^2$

Subtract the common terms.

$- 2uvxy \le u^2y^2 + x^2v^2$

At which point we could invoke the AM-GM inequality. Or we could say.

$0 \le u^2y^2 + 2uvxy + x^2v^2$

Which is a perfect square.

$0 \le (uy + xv)^2$

And square numbers are always positive.

Our proposition is true.

Answer 3

Let $\vec{A}= <u,v>$ and $\vec{B}= <x,-y>$

According to the Cauchy- Schwartz inequality we have $$|\vec A.\vec B| \le ||\vec A||.||\vec B||$$

Note that $\vec A.\vec B=ux-vy$ and $$ ||\vec A||. ||\vec B||=\sqrt {(a^2+b^2)(u^2+v^2)}$$ Thus $$|ux-vy|\le \sqrt {(a^2+b^2)(u^2+v^2)}$$