Prove the Cauchy-Schwartz inequality |(f,g)|<=||f||.||g|| by writing g=af+h with

Question

Prove the Cauchy-Schwartz inequality |(f,g)|<=||f||.||g||
by writing g=af+h with h being independent of f and noting that
||g||^2=||af||^2+||h||^2>=||af||^2

Explanation / Answer

: For real a and b and for x > 0, and y > 0, one has (a + b) 2 x + y = a 2 x + b 2 y . (1) Naturally, this lemma is trivial — once it is conceived. Just by expansion and factorization one ?nds that it is simply a restatement of (ay - bx) 2 = 0. What I ?nd attractive about this inequality is the way that it is self-generalizing. For example, if we replace b by b + c and replace y by y + z the we ?nd (a + b + c) 2 x + y + z = a 2 x + (b + c) 2 y + z = a 2 x + b 2 y + c 2 z . (2) Repeating this step then gives us the general relationship (a1 + a2 + · · · + an) 2 x1 + x2 + · · · + xn = a 2 1 x1 + a 2 2 x2 + · · · + a 2 n xn , (3) from which we can see Cauchy-Schwarz inequality at a glance. One simply sets ak = akßk and xk = ß 2 k .

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