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## experimentX if $$a$$ and $$b$$ are two vectors in linear vector space, prove Cauchy-Swartz inequality $$|a|.|b| \geq |a.b|$$ one year ago one year ago

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1. phi

I have seen a proof that u|dw:1346599918682:dw|ses the idea that a quadratic y= a x^2 + bx +c is always positive for all values of x if its discriminate is negative: this means complex roots

2. experimentX

I'm particularly looking for something to write in exam

3. phi

use this idea with | a + x b|^2 where a and b are vectors and x is a scalar this must always be positive for all values of x

4. phi

write the magnitude squared as a dot product | a + x b|^2 ≥0 $|a|^2 + 2(a \cdot b) x + |b|^2 x^2 ≥ 0$ as noted above, the discriminate of this quadratic in x must be negative to guarantee all values ≥ 0 for all x: $4 (a \cdot b)^2 -4 |a|^2 |b|^2 ≤ 0$

5. experimentX

how do you generalize it to n space?

6. phi

The vectors a and b are in n space.

7. experimentX

i see..