Proving Inequalities with Cauchy

In the Chinese book: 用柯西不等式证明不等式

Build two lists whose pairwise products lose every variable, leaving a constant.

📖 §4.1 of 《平均值不等式与柯西不等式》M18 Construction line

What this move does

The Cauchy inequality $(\sum u_{i}^{2}) (\sum v_{i}^{2}) \geq (\sum u_{i} v_{i})^{2}$ is not hard in itself. The hard part is seeing where the two lists are. The move:

Find two factors in the problem that can play the role of “sums of squares” (often two brackets multiplied together, or a sum you are given).
Split each term into $u_{i}^{2}$ and $v_{i}^{2}$ so that in the products $u_{i} v_{i}$ the variables cancel, leaving only constants or known quantities.
Apply Cauchy, then check whether the equality condition (all ratios $u_{i} : v_{i}$ equal) can be met under the constraints.

On the workbench above, $u_{i} = p_{i} / x_{i}$ and $v_{i} = x_{i}$ : their sums of squares are $\sum \frac{p _{i}}{x _{i}}$ and $\sum x_{i}$ , while the product $u_{i} v_{i} = p_{i}$ does not depend on $x_{i}$ . When the variables vanish from the products, the lower bound is a constant.

Predict first

$a, b, c > 0$ . What is the minimum of $(\frac{1}{a} + \frac{9}{b} + \frac{16}{c}) (a + b + c)$ ?

26

36

64

81

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

Work through one

Example: prove (1/a + 9/b + 16/c)(a + b + c) ≥ 64

Step 1 / 5

Spot the two sums of squares
The left side is two brackets multiplied together, exactly the shape of the right side of Cauchy, $\sum u_{i}^{2} \cdot \sum v_{i}^{2}$ . Try $a = b = c = 1$ first: the left side is $26 \times 3 = 78$ .

$(\frac{1}{a} + \frac{9}{b} + \frac{16}{c}) (a + b + c)$
Split into u² and v²
The variables vanish from the products
Equality: every pair of bars has the same shape
The general pattern

Another split: let 1 be the second list

Not every problem has two brackets. When there is only one sum, the other list is often just $1, 1, \dots, 1$ .

Example: a + b + c = 1, a, b, c > 0; prove √(3a+1) + √(3b+1) + √(3c+1) ≤ 3√2

Step 1 / 3

We want an upper bound for a sum of products
The left side is a sum of three square roots; read it as $\sum u_{i} v_{i}$ with $u_{i} = 1$ and $v_{i} = 3 a_{i} + 1$ . The picture uses some arbitrary $v_{i}$ .

$1 \cdot 3 a + 1 + 1 \cdot 3 b + 1 + 1 \cdot 3 c + 1$
Both sums of squares are known
Equality: all the v have the same length

In the picture the $v_{i}$ can only be multiples of $0.5$ , so the third step uses $1.5$ instead of $2$ . What matters is that they are all equal: the slack is $0$ .

Practice

Exercise 1Basic

$a, b > 0$ . Find the minimum of $(a + b) (\frac{1}{a} + \frac{4}{b})$ .

Exercise 2Basic

Positive numbers $x, y, z$ satisfy $x + y + z = 1$ . Find the maximum of $(x + 2 y + 3 z)^{2}$ .

Exercise 3Advanced

Let $x, y, z > 0$ . Prove that $(x + y + z) (\frac{1}{x + y} + \frac{1}{y + z} + \frac{1}{z + x}) \geq \frac{9}{2}$ .

Exercise 4Challenge

Let $a, b, c > 0$ with $ab c = 1$ . Prove that $\frac{1}{a} + \frac{4}{b} + \frac{9}{c} \geq \frac{36}{a + b + c}$ , and explain why this does not directly give $\frac{1}{a} + \frac{4}{b} + \frac{9}{c} \geq 12$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Carrying on while the products still contain variables

Given

a + b + c = 3

, prove

(a^{2} + b^{2} + c^{2}) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9

: by Cauchy, the left side is

\geq (a + b + c)^{2} \geq 9

What went wrong? Think first, then open

Diagnosis The first step is fine, but the right side $(a + b + c)^{2}$ still contains variables. And it is in fact at most $3 (a + b + c) = 9$ (Cauchy again), so the second step goes the wrong way.

Here bound each factor on its own:

a^{2} + b^{2} + c^{2} \geq \frac{( a + b + c ) ^{2}}{3} = 3

and

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{9}{a + b + c} = 3

; multiplying gives

\geq 9

“A good construction makes the products constant; if variables are left, check the direction first”

Exhibit

A lower bound that is never reached

a \geq 2

b > 0

a + b = 4

, so

\frac{1}{a} + \frac{9}{b} \geq \frac{( 1 + 3 ) ^{2}}{a + b} = 4

, hence the minimum is

4

What went wrong? Think first, then open

Diagnosis Equality needs $b = 3 a$ , i.e. $a = 1$ , but the problem requires $a \geq 2$ . The inequality itself is true; the value $4$ simply can never be reached.

[2, 4)

the function

f (a) = \frac{1}{a} + \frac{9}{4 - a}

is increasing, so its minimum is at

a = 2

\frac{1}{2} + \frac{9}{2} = 5

“Before you write “the minimum is”, put the equality condition back into the constraints”