Cauchy Schwarz Inequality for numbers

The CS inequality is given by

x1y1+x2y2x21+x22y21+y22

I read that if x1=cy1 and x2=cy2, then equality holds.

But I reduced the above to cc2=|c|. So isn’t this only true if c>0?

Answer

Assume x1,,y2 are real numbers. We have (x21+x22)(y21+y22)(x1y1+x2y2)2=(x1y2x2y1)2. This is the case n=2 of an identity due to Lagrange.

(See more on this, positive polynomials, sums of squares of polynomials, and Hilbert’s 17th problem, in this old blog post of mine.)

This means that |x1y1+x2y2|x21+y21y21+y22, with equality iff det that is, iff either there is a constant c such that x_1=cy_1 and x_2=cy_2, or else y_1=y_2=0.

Since a\le|a| for any real number a, it follows that x_1y_1 + x_2y_2 \leq \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}, with equality as above, except that now we further need c\ge 0.

Attribution
Source : Link , Question Author : Hawk , Answer Author : Andrés E. Caicedo

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