would this be a correct proof of the cauchy-schwartz inequality:
abs=absolute value
abs(u*v) is less than or equal to abs(u)*abs(v).
Then you divide both sides by abs(u)*abs(v) so that you get cos(theta) is less than or equal to 1.
If you know that the dot product of vectors u and v is |u| |v| cos theta
then you can surely say that
u dot v </= |u| |v|
No, the proof you provided is not a correct proof of the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for any two vectors u and v in a vector space with an inner product, the absolute value of their inner product is less than or equal to the product of their norms (absolute values). Mathematically, it can be written as:
|u·v| ≤ ||u|| * ||v||
To prove this inequality, you can follow the steps below:
1. Start by considering the following quadratic expression:
0 ≤ ||u - λv||^2, where λ is a scalar.
2. Expand the expression using the properties of the inner product:
0 ≤ (u - λv) · (u - λv)
0 ≤ u · u - 2λ(u · v) + λ^2(v · v)
0 ≤ ||u||^2 - 2λ(u · v) + λ^2||v||^2
3. Notice that this quadratic expression is always non-negative, so its discriminant (b^2 - 4ac) must be non-positive:
4(u · v)^2 - 4||u||^2 ||v||^2 ≤ 0
4. Simplify the inequality:
(u · v)^2 ≤ ||u||^2 ||v||^2
5. Taking the square root of both sides (since both sides are non-negative), we obtain the Cauchy-Schwarz inequality:
|u · v| ≤ ||u|| ||v||
Notice that in the correct proof, there is no division by absolute values or cos(theta). The inequality is derived purely from the properties of the inner product and the non-negativity of the quadratic expression.