The Cauchy–Schwarz Inequality

A dot product and an angle

Two vectors a and b drawn from one point meet at an angle $\theta$θ, and their dot product is $a\cdot b=|a||b|\cos \theta$a · b = |a| |b| cos θ. In components it is also $a_1{b}_1+a_2{b}_2$a₁b₁ + a₂b₂, the first components multiplied plus the second components multiplied.

Take a = (4, 0) and b = (3, 3). The components give a · b = 4 × 3 + 0 × 3 = 12. The lengths are |a| = 4 and $|b|=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$|b| = √(3² + 3²) = √18 = 3√2, so $|a||b|=12\sqrt{2}$|a| |b| = 12√2, which is about 16.97.

So $12\sqrt{2}\cos \theta =12$12√2 cos θ = 12, which gives $\cos \theta =\frac{1}{\sqrt{2}}$cos θ = 1/√2 and $\theta =45^{\circ }$θ = 45°. The dot product, 12, is smaller than the product of the lengths, 16.97, because cos 45° is less than 1.

a = (4, 0) and b = (3, 3) meet at 45°. Their dot product is 12, and the product of their lengths is $4\times 3\sqrt{2}=12\sqrt{2}$4 × 3√2 = 12√2, about 16.97.

The projection

Drop a perpendicular from the head of b onto the line of a. The piece of that line from the tail to the foot of the perpendicular is the projection of b on a, and its length is $|b|\cos \theta$|b| cos θ. The dot product is the length of a times that projection.

For a = (4, 0) and b = (3, 3), the perpendicular from (3, 3) lands at (3, 0), so the projection is 3. Then |a| × 3 = 4 × 3 = 12, the same dot product as before.

A projection is never longer than the vector it comes from: it is one side of a right triangle whose hypotenuse is b. Here the projection is 3 and |b| is about 4.24.

The projection of b = (3, 3) on a = (4, 0), in gold, has length 3, and 4 × 3 = 12 = a · b.

The inequality

A cosine lies between −1 and 1, so $|\cos \theta |\le 1$|cos θ| ≤ 1. Multiply both sides by the lengths |a| |b|, which are not negative: $|a||b||\cos \theta |\le |a||b|$|a| |b| |cos θ| ≤ |a| |b|. The left side is the size of the dot product, so $|a\cdot b|\le |a||b|$|a · b| ≤ |a| |b|. That is the Cauchy–Schwarz inequality.

Written in components and squared, it reads ${\left(a_1{b}_1+a_2{b}_2\right)}^2\le \left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)$(a₁b₁ + a₂b₂)² ≤ (a₁² + a₂²)(b₁² + b₂²). For a = (4, 0) and b = (3, 3) the left side is $12^2=144$12² = 144 and the right side is 16 × 18 = 288.

The bound is a product of the lengths, not a sum. Two vectors of lengths 5 and 5 have a dot product of at most 5 × 5 = 25, and never 10.

a has length 4 and b has length 3, so the bound is 4 × 3 = 12. At $\theta =60^{\circ }$θ = 60° the projection is 1.5 and a · b = 6. Swing b round: the dot product reaches 12 only at 0°, and its size reaches 12 again at 180°, where it is −12.

A proof with no angle

The components give a second proof, which needs no angle at all. Multiply out the right side and take away the left side: $\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)-{\left(a_1{b}_1+a_2{b}_2\right)}^2=a_1^2{b}_2^2-2a_1{b}_2{a}_2{b}_1+a_2^2{b}_1^2$(a₁² + a₂²)(b₁² + b₂²) − (a₁b₁ + a₂b₂)² = a₁²b₂² − 2a₁b₂a₂b₁ + a₂²b₁², because the terms $a_1^2{b}_1^2$a₁²b₁² and $a_2^2{b}_2^2$a₂²b₂² cancel.

What is left is a perfect square, ${\left(a_1{b}_2-a_2{b}_1\right)}^2$(a₁b₂ − a₂b₁)². A square is never negative, so the right side is at least the left side, and the inequality holds for any four numbers $a_1$a₁, $a_2$a₂, $b_1$b₁ and $b_2$b₂.

Check it on a = (4, 0) and b = (3, 3): 288 − 144 = 144, and ${\left(4\times 3-0\times 3\right)}^2=12^2=144$(4 × 3 − 0 × 3)² = 12² = 144.

When the two sides are equal

Equality needs $|\cos \theta |=1$|cos θ| = 1, so $\theta =0^{\circ }$θ = 0° or $\theta =180^{\circ }$θ = 180°: the two vectors lie along one line, pointing the same way or opposite ways. That is what parallel means here. The projection of b is then the whole of b.

The proof in components says the same. Equality needs ${\left(a_1{b}_2-a_2{b}_1\right)}^2=0$(a₁b₂ − a₂b₁)² = 0, so $a_1{b}_2=a_2{b}_1$a₁b₂ = a₂b₁, which says that one vector is a number times the other.

For a = (4, 2) and b = (2, 1), a is 2 times b. Then a · b = 8 + 2 = 10, and $|a||b|=\sqrt{20}\times \sqrt{5}=\sqrt{100}=10$|a| |b| = √20 × √5 = √100 = 10. For a = (3, 4) and c = (−6, −8), c is −2 times a: a · c = −18 − 32 = −50, and |a| |c| = 5 × 10 = 50, so |a · c| = 50, equal to the bound.

The equality case is how the inequality finds a largest value. Among all vectors b of length 5, the dot product (3, 4) · b is at most 5 × 5 = 25, and it reaches 25 only when b points along (3, 4), at b = (3, 4) itself: 9 + 16 = 25.

a = (4, 2) and b = (2, 1) lie along one line, and a · b = 10 = |a| |b|. Against −b = (−2, −1), pointing the other way, a · (−b) = −10, and its size is again the product of the lengths.

Three components, and the triangle inequality

The inequality holds with three components, or with any number: ${\left(a_1{b}_1+a_2{b}_2+a_3{b}_3\right)}^2\le \left(a_1^2+a_2^2+a_3^2\right)\left(b_1^2+b_2^2+b_3^2\right)$(a₁b₁ + a₂b₂ + a₃b₃)² ≤ (a₁² + a₂² + a₃²)(b₁² + b₂² + b₃²). For (1, 2, 2) and (2, 3, 6), the dot product is 2 + 6 + 12 = 20, and the lengths are $\sqrt{9}=3$√9 = 3 and $\sqrt{49}=7$√49 = 7, so $20\le 3\times 7=21$20 ≤ 3 × 7 = 21.

It also proves the triangle inequality for vectors. Expand $|a+b{|}^2=|a{|}^2+2a\cdot b+|b{|}^2$|a + b|² = |a|² + 2a · b + |b|². The middle term is at most 2|a| |b|, so $|a+b{|}^2\le |a{|}^2+2|a||b|+|b{|}^2={\left(|a|+|b|\right)}^2$|a + b|² ≤ |a|² + 2|a| |b| + |b|² = (|a| + |b|)², and taking square roots gives $|a+b|\le |a|+|b|$|a + b| ≤ |a| + |b|.

The usual mistakes

Dropping the size bars. A dot product can be negative, and a · c = −50 is at most 50 without saying much. The inequality bounds the size of the dot product, |a · b|, and a negative dot product reaches the bound when the vectors point opposite ways.

Taking perpendicular vectors for the equality case. Perpendicular vectors have $\cos \theta =0$cos θ = 0, so their dot product is 0, as far from the bound as it can be.

Adding the lengths. The bound for lengths 4 and 3 is 4 × 3 = 12, not 4 + 3 = 7.

Matching the wrong things in an application. In the advertising problem below, the weights (2, 3, 6) are matched with the square roots of the spends, so equality puts the spends in the ratio 4 : 9 : 36, not 2 : 3 : 6.

Budgets and cables

In the first application, the new customers from three advertising channels are the dot product of the weights (2, 3, 6) with the square roots of the three spends. The weights have length 7 and the square roots of the spends have length $\sqrt{4900}=70$√4900 = 70, so the customers number at most 7 × 70 = 490.

In the second, the total current of 30 A is a dot product, and the heat lost sits inside the length of one of the two vectors, so the inequality gives the least heat instead of the most.