Is there is geometric proof of the Cauchy-Schwarz Inequality with 2 terms (not the vector one)?
$$(a² + b²)(c² + d²) ≥ (ac + bd)² $$
A geometric representation of this identity would also mean the same:
$$(a² + b²)(c² + d²) = (ac + bd)² + (bc – ad)²$$
Thanks in advance!
Best Answer
I guess this is the best I can squeeze out of my brain right now:
The parallelogram inside of the rectangle has side lengths $\sqrt{a^2+b^2}$ and $\sqrt{c^2+d^2}$, and so, its area is at most
$$\sqrt{(a^2+b^2)(c^2+d^2)}.$$
On the other hand, its area can be computed by
$$(b+c)(a+d)-ab-cd=ac+bd.$$