Cauchy-Schwarz Inequality by sum of squares.
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Apparently your teacher wants you to use Lagrange's identity:
$$||a^2||\cdot||b^2|| - (a \cdot b)^2 = \sum_{i = 1}^{n - 1} \sum_{j = i + 1}^n (a_i b_j - a_j b_i)^2$$
which is indeed a sum of squares, thus it is positive, therefore $||a^2||\cdot||b^2|| \geqslant (a \cdot b)^2$ for all vectors $a$, $b$.
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Jerome Turner
Updated on March 17, 2020Comments
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Jerome Turner over 3 years
i'm trying to solve: Prove the Cauchy-Schwarz inequality by writing $||x||^2||y||^2− |⟨x,y⟩|^2$ as a sum of squares.
I'm fairly well versed in Cauchy Schwarz and know several proofs but I'm confused what it means to be "writing $||x||^2||y||^2− |⟨x,y⟩|^2$ as a sum of squares"
I realize $||x||^2||y||^2− |⟨x,y⟩|^2= (x \cdot x)(y\cdot y)-(x\cdot y)^2 $ but how do I apply to prove?Any help appreciated.