
#1
Sep206, 11:13 PM

P: 1,239

Lets say we have: [tex] (a_{1}b_{1} + a_{2}b_{2} + ... + a_{n}b_{n})^{2} \leq (a_{1}^{2} + a_{2}^{2} + ... + a_{n}^{2})(b_{1}^{2} + b_{2}^{2} + ... + b_{n}^{2}) [/tex].
Let [tex] A = a_{1}^{2} + a_{2}^{2} + ... + a_{n}^{2} , B = a_{1}b_{1} + a_{2}b_{2} + ... + a_{n}b_{n}, C = b_{1}^{2} + b_{2}^{2} + ... + b_{n}^{2} [/tex]. Thus we have [tex] AC \geq B^{2} [/tex]. From [tex] 0\leq (a_{1} + tb_{1})^{2} + (a_{2} + tb_{2})^{2} + ... + (a_{n} + tb_{n})^{2} [/tex] where [tex] t [/tex] is any real number, we obtain [tex] 0 \leq A + 2Bt + Ct^{2} [/tex]. Completing the square, we obtain [tex] Ct^{2} + 2Bt + A = C(t + \frac{B}{C})^{2} + (A  \frac{B^{2}}{C}) [/tex]. From this step, how do we obtain [tex] 0 \leq A  \frac{2B^{2}}{C} + \frac{B^{2}}{C} = \frac{ACB^{2}}{C} [/tex], implying that [tex] AC  B^{2} \geq 0 [/tex]? Thanks 



#2
Sep306, 12:34 AM

P: 3,177

you need to show that for the minimal value of t the inequality still holds.
from algebra we know that for C>=0 t=B/C. 



#3
Sep306, 08:59 AM

P: 113

Hi courtrigrad,
Why complete the square? The graph of [itex] Ct^2 + 2Bt + A[/itex] is that of a porabola opening upwards. Since this quadratic equation is greater than or equal to zero for all values of t there can either be a single root of multiplicity 2, or none at all. What does this tell you about the discriminant of the equation? 



#4
Sep306, 09:32 AM

P: 1,239

Cauchy Schwarz Inequality
yeah, but I am referring this out of Courant's book. Just trying to see how he approached it. [tex] B^{2}  4AC \leq 0 [/tex]
Thanks 



#5
Sep306, 10:09 AM

P: 113

Probably a typographical error, but the discriminant should be 4B^2  4AC.
Perhaps Courant is implying you plug in the value t = B/C, in which case you would get [tex]A  \frac{B^2}{C} \geq 0 [/tex] or if you plug it into the original quadratic you would get the equvalent [tex]0 \leq A  \frac{2B^2}{C} + \frac{B^2}{C} [/tex] as he did. He probably completes the square to motivate this choice of t. 


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