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This is the proof of Cauchy-Schwarz inequality given in "One variable calculus with introduction to linear algebra" by Tom M. Apostol.

In last paragraph It is said that the right hand side has smallest value when x=... etc!

so we proved.

My question is why we choose smallest value? why not largest value? are we allowed to choose like this? Is this logically correct to choose like this? we choose some particular x, doesn't it went wrong if we choose like this? please explain this.

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We have $$A\left(x+\frac{B}{A}\right)^2+\frac{AC-B^2}{A}$$. $A$ is a sum of squares, it can never be negative, and we want $$Ax^2+2Bx+C\geq 0$$. The first summand is $$A\left(x+\frac {B}{A}\right)\geq 0$$ for all real $x$, then our whole sum is not negative if also the second summand is non negative if $$AC-B^2\geq 0$$

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  • $\begingroup$ if first summand is non negative .Did we still need $AC-B^2>=0$? $\endgroup$ – Sathasivam K Apr 30 '18 at 13:20
  • $\begingroup$ Yes that is what we Need, this is the Cauchy Schwarz inequality, but if the first summand is strictly positive it Can be that the whole sum is not negative, and we do not Need $$AC-B^2\geq 0$$ so we use $$x=-\frac{B}{A}$$ and our first summand is Zero and it MUST be $$AC-B^2\geq 0$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 30 '18 at 13:22
  • $\begingroup$ Thanks a lot sir! $\endgroup$ – Sathasivam K Apr 30 '18 at 13:41
  • $\begingroup$ please forgive me for asking one more questions! doesn't it looks like a particular case ?if $x\neq \cfrac{-B}{A}$ does this inequality still make sense ? $\endgroup$ – Sathasivam K Apr 30 '18 at 13:45
  • $\begingroup$ If $$x\neq -\frac{B}{A}$$ it Can be that $$A\left(x+\frac{B}{A}\right)^2+\frac{AC-B^2}{A}\geq 0$$ without the condition $$AC-B^2\geq 0$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 30 '18 at 13:48

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