# Understanding Cauchy-Schwarz Inequality

• I
• Mr Davis 97
In summary, the Cauchy-Schwarz inequality can be used to bound the maximum and minimum values of a function. In the specific case of ##f(x,y) = 2\sin x \sin y + 3\sin x \cos y + 6 \cos x##, the inequality gives the bound of ##-7 \le f(x,y) \le 7##. However, it is possible that the maximum and minimum values could be smaller than this bound and the inequality is strict. This can be seen in the example of ##f(x,x)##, where the maximum value is approximately 7 but the vectors used in the Cauchy-Schwarz inequality are not multiples of each other, leading
Mr Davis 97
I am trying to find the max and min values of the function ##f(x,y) = 2\sin x \sin y + 3\sin x \cos y + 6 \cos x##. By the Cauchy-Schwarz inequality, we have that ##|f(x,y)|^2 \le (4+9+36) (\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 x) = 49##. Hence ##-7 \le f(x,y) \le 7##.

My question has to do with the last inequality. What information does this inequality convey exactly? Does it say that f(x,y) actually takes on all values and only the values in the interval ##[-7,7]##,and hence -7 and 7 are the min and max? Or does it say -7 and 7 are bounds on the range of f(x,y), and the max and min could actually be smaller values contained in the interval?

Last edited:
In general you are just bounding the maximum and minimum values here, and bound is loose unless you can find a case where one thing is a scalar multiple of another, i.e.

##\big \vert \langle \mathbf a, \mathbf b \rangle \big \vert \leq \big \vert \mathbf a\big \vert_2 \vert \mathbf b\big \vert_2 ##,

with equality iff ##\mathbf a \propto \mathbf b##

(and there is the special case of one of the vectors being the zero vector).
--------
It isn't clear to me why you have

##(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1##

I.e. from what you've given ##x## and ##y## are independent variables. If there is some constraint that x = y, then it holds. Otherwise I can easily come up with a case where it equals 2.

FactChecker
StoneTemplePython said:
In general you are just bounding the maximum and minimum values here, and bound is loose unless you can find a case where one thing is a scalar multiple of another, i.e. ##\big \vert \langle \mathbf a, \mathbf b \rangle \big \vert \leq \big \vert \mathbf a\big \vert_2 \vert \mathbf b\big \vert_2 ##, with equality iff ##\mathbf a \propto \mathbf b## (and there is the special case of one of the vectors being the zero vector).
--------
It isn't clear to me why you have

##(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1##

I.e. from what you've given ##x## and ##y## are independent variables. If there is some constraint that x = y, then it holds. Otherwise I can easily come up with a case where it equals 2.
I meant for that y to be an x, so it is 1.

But does this mean that -7 and 7 are not the min and the max?

Mr Davis 97 said:
I meant for that y to be an x, so it is 1.

I don't know what this means. You explicitly defined a two variable function ##f(x,y)##. But I'll assume the identity holds and move on:

Mr Davis 97 said:
But does this mean that -7 and 7 are not the min and the max?
##\mathbf a = \begin{bmatrix}
2\\
3\\
6
\end{bmatrix}##

##\mathbf b = \begin{bmatrix}
\sin x \sin y\\
\sin x \cos y \\
\cos x
\end{bmatrix}##

is there some ##\gamma##, where ##\mathbf a = \gamma \mathbf b##?

If not then Cauchy's inequality is strict and you can be certain the function never takes on the values of -7 and 7.

Last edited:
StoneTemplePython said:
It isn't clear to me why you have

##(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1##
Especially since the OP has concluded that |f|≤7 but f(π/2, 0) = 9.

StoneTemplePython
Mr Davis 97 said:
I meant for that y to be an x, so it is 1.
Do you mean in the original problem or just in this part? f(π/2, 0) = 9, so if you need to correct the original problem, you should do that and see what changes.

StoneTemplePython said:
##\mathbf a = \begin{bmatrix}
2\\
3\\
6
\end{bmatrix}##

##\mathbf b = \begin{bmatrix}
\sin x \sin y\\
\sin x \cos y \\
\cos y
\end{bmatrix}##

is there some ##\gamma##, where ##\mathbf a = \gamma \mathbf b##?

If not then Cauchy's inequality is strict and you can be certain the function never takes on the values of -7 and 7.
It's interesting to note that at about x=0.550818287203578, the function f(x,x) is about maximized at about 7 and the vectors you give above are essentially multiples of each other.

FactChecker said:
It's interesting to note that at about x=0.550818287203578, the function f(x,x) is about maximized at about 7 and the vectors you give above are essentially multiples of each other.

yea -- by my calculations

- - - -

edit:

we have

##\gamma \cos(x) = 6##
## 6 \sin(x) = \big(\gamma \cos(x)\big) \sin(x) =\gamma \sin(x) \cos(x) = 3##
hence we have ##\sin(x) = \frac{1}{2}##

Finally

##\gamma \frac{1}{2}\frac{1}{2} = \gamma \sin(x) \sin(x) = 2##
hence ##\gamma = 8##

unfortunately this implies

##8 \cos(x) = 6##, or ##\cos(x) = 0.75##, while in fact we have something more like ##\cos(x) = 0.87##, so the ##\gamma ## doesn't quite exist and the inequality is strict. Nevertheless, interesting to get so close to 7.

Last edited:

## What is the Cauchy-Schwarz Inequality?

The Cauchy-Schwarz Inequality is a fundamental mathematical concept that states that for any two vectors in a vector space, the dot product of the two vectors is less than or equal to the product of the magnitudes of the vectors. This can also be written as |x · y| ≤ ||x|| ||y||, where x and y are vectors and ||x|| and ||y|| are their respective magnitudes.

## Why is the Cauchy-Schwarz Inequality important?

The Cauchy-Schwarz Inequality has many important applications in mathematics, physics, and engineering. It is used in a variety of mathematical proofs and calculations, and it is also a key concept in understanding vector spaces, inner product spaces, and metric spaces. Additionally, it has practical applications in fields such as signal processing, statistics, and machine learning.

## How is the Cauchy-Schwarz Inequality proven?

The Cauchy-Schwarz Inequality can be proven using several different methods, including the Cauchy-Bunyakovsky-Schwarz (CBS) Inequality, the Holder's Inequality, and the Minkowski Inequality. Each of these proofs uses a different approach, but they all rely on the basic idea that the dot product of two vectors is related to the magnitudes of the vectors and the angle between them.

## Can the Cauchy-Schwarz Inequality be applied to more than two vectors?

Yes, the Cauchy-Schwarz Inequality can be extended to apply to more than two vectors. This is known as the generalized Cauchy-Schwarz Inequality, and it states that for any n vectors in a vector space, the dot product of any two of those vectors is less than or equal to the product of the magnitudes of all n vectors. This concept is often used in higher-level mathematics and physics.

## Are there any limitations to the Cauchy-Schwarz Inequality?

While the Cauchy-Schwarz Inequality is a powerful and widely applicable concept, it does have some limitations. One limitation is that it only applies to finite-dimensional vector spaces. Additionally, it is not always possible to achieve equality in the inequality, and there are certain situations where the inequality may not hold. These limitations should be considered when applying the Cauchy-Schwarz Inequality in practical situations.

• General Math
Replies
8
Views
1K
• General Math
Replies
1
Views
954
• General Math
Replies
4
Views
1K
• General Math
Replies
2
Views
997
• General Math
Replies
1
Views
2K
• Calculus
Replies
3
Views
2K
• Precalculus Mathematics Homework Help
Replies
5
Views
836
• Topology and Analysis
Replies
4
Views
501
• Calculus and Beyond Homework Help
Replies
4
Views
995
• Linear and Abstract Algebra
Replies
2
Views
1K