The Cauchy Inequality and Its Proofs

Try this

Drag $v$ onto the same line as $u$ (same direction or opposite, either works). What does “Gap = area²” become?
Keep $u$ fixed and swing $v$ once around the origin (keeping its length roughly the same). When is the left side $(u \cdot v)^{2}$ largest? When is it $0$ ?
Look at the thick segment: the projection of $v$ onto the direction of $u$ . Why is it never longer than $∣ v ∣$ ?
Put the vectors anywhere, compute $(a_{1} b_{2} - a_{2} b_{1})^{2}$ , and compare it with right side minus left side.

For two vectors $u = (a_{1}, a_{2})$ and $v = (b_{1}, b_{2})$ , the dot product is $a_{1} b_{1} + a_{2} b_{2} = ∣ u ∣∣ v ∣ cos θ$ , and $∣ cos θ ∣ \leq 1$ . Squaring gives

(a_{1} b_{1} + a_{2} b_{2})^{2} \leq (a_{1}^{2} + a_{2}^{2}) (b_{1}^{2} + b_{2}^{2}) .

Replace two numbers by $n$ numbers and you get the Cauchy inequality: for any real numbers $a_{1}, \dots, a_{n}$ and $b_{1}, \dots, b_{n}$ ,

(i = 1 \sum n a_{i} b_{i})^{2} \leq (i = 1 \sum n a_{i}^{2}) (i = 1 \sum n b_{i}^{2}),

with equality if and only if the two lists are proportional (there is a $λ$ with $b_{i} = λ a_{i}$ for every $i$ , or all the $a_{i}$ are $0$ ).

Predict first

$u = (3, 4)$ , and $v$ has length $2$ and any direction. What is the largest possible value of $u \cdot v$ ?

Choose an answer first; then you can see the result.

Why: a parabola that never dips below the axis

The geometric argument only covers $n = 2, 3$ . For any $n$ there is a proof that uses nothing but school algebra: a sum of squares is never negative.

Proving Cauchy with the discriminant

Step 1 / 5

Build a sum of squares
Take a variable $t$ , turn each pair of numbers into $(a_{i} t - b_{i})^{2}$ , and add them all up. The dashed curves are the single terms, the solid curve is their sum. Every term is $\geq 0$ , so $f (t) \geq 0$ for every $t$ .

$f (t) = i = 1 \sum n (a_{i} t - b_{i})^{2} \geq 0$
Expand into a quadratic
Never below the axis ⇒ discriminant not positive
When does it touch the t-axis?
The converse holds too

How big is the gap? Lagrange's identity

The discriminant shows the inequality never goes the other way, but it does not say how much is lost. For $n = 2$ , just expand:

Right side minus left side is exactly a square

Step 1 / 5

Numbers first
$u = (3, 1)$ , $v = (1, 2)$ : the left side is $(3 + 2)^{2} = 25$ , the right side $10 \times 5 = 50$ , a gap of $25$ .

$(3 \cdot 1 + 1 \cdot 2)^{2} = 25 \leq 50 = (3^{2} + 1^{2}) (1^{2} + 2^{2})$
Expand the right side
What is left is a perfect square
Flatten it
The same for n numbers

Three proofs, three ways of looking: the angle ( $∣ cos θ ∣ \leq 1$ ), the discriminant (the parabola never dips below the axis) and completing the square (the gap is a sum of squares). They all say the same thing: how tight the inequality is depends entirely on how far the two lists are from being proportional.

Practice

Exercise 1Basic

Real numbers $x, y$ satisfy $3 x + 4 y = 10$ . Find the minimum of $x^{2} + y^{2}$ .

Exercise 2Basic

Real numbers $x, y, z$ satisfy $x^{2} + y^{2} + z^{2} = 4$ . Find the maximum of $x + 2 y + 2 z$ .

Exercise 3Advanced

Real numbers $x, y$ satisfy $x^{2} + 4 y^{2} = 5$ . Find the maximum of $x + y$ .

Exercise 4Challenge

Let $a, b > 0$ . Prove that $(a^{3} + b^{3}) (a + b) \geq (a^{2} + b^{2})^{2}$ , and say when equality holds.

Try this

Why: a parabola that never dips below the axis

Proving Cauchy with the discriminant

How big is the gap? Lagrange's identity

Right side minus left side is exactly a square

Practice

Mistake Museum · Exhibits from this stop

Writing only one sum of squares

Mistaking “proportional” for “equal”