Valid proof of Cauchy-Schwarz inequality?

In summary, the Cauchy-Schwarz inequality states that for any two vectors u and v, the absolute value of their inner product is less than or equal to the product of their lengths. A student has proposed a simpler proof using the definition of the dot product and the upper and lower bounds of the cosine function. However, the validity of this proof has not been confirmed and it may not be suitable for an exam in a multivariable and complex calculus course. Alternatively, a longer proof using the properties of the dot product can be used.
  • #1
phosgene
146
1

Homework Statement



I was discussing the proof for the Cauchy-Schwarz inequality used in our lectures, and another student suggested an easier way of doing it. It's really, really simple. But I haven't seen it anywhere online or in textbooks, so I'm wondering if it's either wrong or is only valid in certain scenarios.

The quick version is this:

[itex]u \bullet v = ||u|| ||v|| cosθ[/itex] where θ is between 0 and [itex]\pi[/itex].

The upper and lower bounds of |cosθ| are 1 and 0, respectively, so [itex]|u \bullet v | \leq ||u|| ||v||[/itex]
 
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  • #2
phosgene said:

Homework Statement



I was discussing the proof for the Cauchy-Schwarz inequality used in our lectures, and another student suggested an easier way of doing it. It's really, really simple. But I haven't seen it anywhere online or in textbooks, so I'm wondering if it's either wrong or is only valid in certain scenarios.

The quick version is this:

[itex]u \bullet v = ||u|| ||v|| cosθ[/itex] where θ is between 0 and [itex]\pi[/itex].

The upper and lower bounds of |cosθ| are 1 and 0, respectively, so [itex]|u \bullet v | \leq ||u|| ||v||[/itex]

You need to tell us what is the definition of inner product that you start with. If it is ##u \cdot v = \sum_i u_i v_i,## then you first need to prove that ##u \cdot v = |u| |v| \cos \theta.## Actually, it is sometimes done in the opposite way: from ##u \cdot v = \sum_i u_i v_i,##, one proves the C_S inequality, and then notes that one can *define* ##\theta## from
[tex]\cos \theta =\frac{u \cdot v}{|u| |v|}.[/tex]
 
  • #3
Thanks for the reply, so it seems that the proof in my original post would probably not be a good one to use in an exam? The course in question is multivariable and complex calculus. The proof that I've used myself relies on algebraically manipulating the square of the length of the difference of two vectors ( [itex] ||a \widehat{x} - b \widehat{y} ||^2 \geq 0 [/itex] ) using the properties of the dot product, rather than the cosine argument. It's the proof I've seen in the textbooks I have, but it's a bit longer to write out.
 

Related to Valid proof of Cauchy-Schwarz inequality?

1. What is the Cauchy-Schwarz inequality?

The Cauchy-Schwarz inequality, also known as the Cauchy-Bunyakovsky-Schwarz inequality, is a fundamental inequality in mathematics that states that the inner product of two vectors is less than or equal to the product of their magnitudes.

2. How is the Cauchy-Schwarz inequality used in mathematics?

The Cauchy-Schwarz inequality is used in various fields of mathematics, including linear algebra, functional analysis, and probability theory. It is a powerful tool for proving other mathematical theorems and inequalities.

3. Can you provide an example of how the Cauchy-Schwarz inequality is applied?

Sure, one example is in the proof of the arithmetic-geometric mean inequality. This inequality states that the arithmetic mean of a set of numbers is always greater than or equal to the geometric mean. The Cauchy-Schwarz inequality is used to prove this by considering the vectors [1,1,...,1] and [x1,x2,...,xn] and their inner product.

4. Is the Cauchy-Schwarz inequality a strict inequality?

No, the Cauchy-Schwarz inequality can be an equality in some cases. For example, if the two vectors are parallel, then their inner product will be equal to the product of their magnitudes, resulting in an equality in the Cauchy-Schwarz inequality.

5. Is the Cauchy-Schwarz inequality only applicable to real numbers?

No, the Cauchy-Schwarz inequality holds for complex numbers as well. In fact, the inequality can be extended to inner product spaces, which include both real and complex numbers.

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