The Cauchy-Schwarz inequality, also known as the Cauchy–Bunyakovsky–Schwarz inequality, states that for all sequences of real numbers and , we have
Equality holds if and only if for a non-zero constant . It can be generalized to Hölder's inequality.
Starting from the definition, we will head into the usage of the Cauchy-Schwarz inequality.
Cauchy-Schwarz inequality states that for all real numbers and , we have
where equality holds if and only if for some constant , for all which have .
There are many reformulations of this inequality. There's also a vector form and a complex number version of it. But we only need the above elementary form to tackle Olympiad problems and problems in other areas.
To get a feel of it, let's consider the case of terms. Let the two sequences be and . Then by Cauchy-Schwarz inequality we have
This easily follows from the Fibonacci-Brahmagupta identity
Notice that the equality holds when if the terms are non-zero. The case of 3 terms states that
Let's work through an example followed by a problem to try yourself.
Given such that , prove that
By Cauchy-Schwarz, we have
We have equality when is a constant for all , which happens when
The following is one of the most common examples of the use of Cauchy-Schwarz. We can easily generalize this approach to show that if , then the maximum value of is .
If , what is the maximum value of
Hence, with equality holding when .
Together with , we get
Making the right choice of and can simplify a problem. The following problem will make you realize it:
For positive reals , show that
At first glance, it is not clear how we can apply Cauchy-Schwarz, as there are no squares that we can use. Furthermore, the RHS is not a perfect square. The power of Cauchy-Schwarz is that it is extremely versatile, and the right choice of and can simplify the problem.
By Cauchy-Schwarz, we have
As such, . Multiplying throughout by , we obtain the given inequality.
Proof of Titu's Lemma
For positive reals and , prove that
It is obtained by applying the substitution and into the Cauchy-Schwarz inequality.
It then becomes
Let's give it one more try by yourself with this practical use of the Cauchy-Schwarz inequality:
An often used consequence of the Cauchy-Schwarz inequality is Titu's lemma. It is so important that we devote an entire page to it. It states that for and the equality similarly holds iff .
On the left diagram above, the red rectangle is inscribed in a larger outside rectangle.
On the right, in the same outside rectangle, two blue rectangles are formed by the perpendicular lines arising from 2 adjacent vertices of the red rectangle.
Which one has a larger area, the red region or the blue region?
There are various ways to prove this inequality. A short proof is given below.
Consider the function
Being a sum of squares, is always non-negative. Now we expand out and collect terms. Then it becomes
which is a quadratic in . The graph of this quadratic is an up parabola. Since , the graph either touches the -axis or stays in the upper half of it. That means either it's a perfect square (has a repeated root) or doesn't have any real root. So the discriminant is . Computing the discriminant, we have
Dividing by and rearranging yields the Cauchy-Schwarz inequality.
Notice that equality holds when has a real root (repeated, of course). From the first form of and using the fact that the sum of squares equals to only when each square equals to , we have for all . That is when is constant for all .
We can also derive the Cauchy-Schwarz inequality from the more general Hölder's inequality. Simply put and , and we arrive at Cauchy Schwarz. As such, we say that Holders inequality generalizes Cauchy-Schwarz.
This section is devoted to defining and proving the vector form of Cauchy-Schwarz inequality.
For all vectors and of a real inner product space,
where is the inner product.
Equality hold if and only if the 2 vectors are linearly dependent.
For a complex vector space, we have
Equality holds if and only if and are linearly dependent, that is, one is a scalar multiple of the other (which includes the case when one or both are zero).
Now we will prove the above claim by the vector form of Cauchy-Schwarz.
If it is clear that we have equality, and in this case and are also trivially linearly dependent. We henceforth assume that is non-zero. We also assume that because otherwise the inequality is obviously true.
From the theory of vectors, we know that is the projection of onto the plane orthogonal to . We will show this is true, by showing that is orthogonal to the vector , or equivalently that . By the linearity of the inner product in its first argument, one has
Since these vectors are orthogonal, we can apply the Pythagorean theorem to
After multiplying by , we obtain the Cauchy–Schwarz inequality.
If the relation in the above expression is actually an equality, then and hence . From the definition of as the projection vector, this implies that there is no part of that is perpendicular to , and hence these two vectors are parallel. This establishes the theorem.
Another approach to prove it, in an algebraic way, is given below.
Consider two sequences and and a matrix, whose composing rule for each element is . For sequences with equal number of terms, we'll have
If we attempt to sum all the terms on the matrix, by the distributive property, this would be given as
Now consider two related sequences and and matrix, whose composing rule for each element is , as follows
Summing all terms gives
Subtracting one sum from the other cancels the diagonal terms and yields
Because for every pair , there exists two terms and , the above difference may be re-written as
Because squares are non-negative, thus , and therefore
The following example is motivated by the vector form of Cauchy-Schwarz.
What is the maximum of the function
Hence, the maximum is , which is achieved when .
In this section, we will learn the applications of Cauchy-Schwarz inequality through some examples and problems.
For , prove that
By Cauchy-Schwarz, we have
It's interesting to know that even triangle inequality in dimensions leads to Cauchy-Schwarz inequality, which can be proved easily.
For real values , show that
Squaring both sides and subtracting common terms, we want to show that
Squaring this and dividing by 4, we get
which is Cauchy-Schwarz inequality and we know its proof.
Here is a simple problem involving the application of Cauchy-Schwarz inequality:
Here are some additional problems you can try to prove on your own:
Prove that if are infinite sequences such that and , then .
[APMO 1991] For positive reals such that , show that
- Show that for , we have
- Prove that
Solve this last problem and you're excellent at Cauchy-Schwarz inequality. Here you go: