Proving Cauchy-Schwarz Inequality Using Completing the Square

[courtrigrad]

Lets say we have: (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}).

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

Thanks

 

[MathematicalPhysicist]

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.

 

[nocturnal]

Hi courtrigrad,

Why complete the square? The graph of Ct^2 + 2Bt + A 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?

 

[courtrigrad]

yeah, but I am referring this out of Courant's book. Just trying to see how he approached it. B^{2} - 4AC \leq 0

Thanks

 

[nocturnal]

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 A - \frac{B^2}{C} \geq 0
or if you plug it into the original quadratic you would get the equvalent 0 \leq A - \frac{-2B^2}{C} + \frac{B^2}{C} as he did.

He probably completes the square to motivate this choice of t.