Proving Cauchy-Schwarz Inequality Using Completing the Square

In summary, we can see that the original inequality can be written as AC \geq B^2, and by completing the square and plugging in the value of t = -B/C, we can obtain the equivalent inequality of 0 \leq A - \frac{2B^2}{C} + \frac{B^2}{C} = \frac{AC-B^2}{C}, which implies that AC - B^2 \geq 0. This shows that the minimal value of t still satisfies the inequality, proving the original statement.
  • #1
courtrigrad
1,236
2
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{AC-B^{2}}{C} [/tex], implying that [tex] AC - B^{2} \geq 0 [/tex]?

Thanks
 
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  • #2
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
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?
 
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  • #4
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
 
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  • #5
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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1. What is the Cauchy Schwarz Inequality?

The Cauchy Schwarz Inequality, also known as the Cauchy-Schwarz-Bunyakovsky Inequality, is a mathematical inequality that relates the dot product of two vectors to their magnitudes. It states that for any two vectors a and b, the absolute value of their dot product is less than or equal to the product of their magnitudes: |ab| ≤ ||a|| ||b||. This inequality has a wide range of applications in mathematics, physics, and engineering.

2. How is the Cauchy Schwarz Inequality derived?

The Cauchy Schwarz Inequality can be derived using the properties of the dot product and the Cauchy-Schwarz Inequality for real numbers. By defining a and b as vectors in n-dimensional space, and using the dot product formula, we can manipulate the terms to arrive at the Cauchy Schwarz Inequality.

3. What are some applications of the Cauchy Schwarz Inequality?

The Cauchy Schwarz Inequality has many applications in mathematics, physics, and engineering. It is used in linear algebra, optimization problems, and probability theory. In physics, it is used to prove the uncertainty principle in quantum mechanics. In engineering, it is used to determine the stability of systems and to optimize signal processing algorithms.

4. Are there any generalizations of the Cauchy Schwarz Inequality?

Yes, there are several generalizations of the Cauchy Schwarz Inequality. The most well-known is the generalized Cauchy Schwarz Inequality, which extends the concept to inner product spaces and complex numbers. Other generalizations include the Hölder's Inequality, Minkowski's Inequality, and the Young's Inequality, which are all based on the same principles as the Cauchy Schwarz Inequality.

5. How is the Cauchy Schwarz Inequality used in optimization problems?

The Cauchy Schwarz Inequality is used in optimization problems to determine the maximum or minimum values of a function. By rewriting the function in terms of vectors and using the Cauchy Schwarz Inequality, we can establish an upper or lower bound for the function. This bound can then be used to find the optimal solution to the optimization problem.

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