The 1821 Cours d’analyse ends with nine notes, a kind of appendix with results Cauchy did not want to put in the main text. The second, on inequalities, contains two gems that every olympiad student knows.

The inequality

Cauchy’s theorem says that, for any real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$,

$$(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2 \le (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2),$$

with equality only when the two lists are proportional.

The proof by Lagrange’s identity

Cauchy started from an identity already found in Lagrange. For two pairs of numbers, just expand:

$$(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1 b_1 + a_2 b_2)^2 = a_1^2 b_2^2 + a_2^2 b_1^2 - 2a_1 b_1 a_2 b_2 = (a_1 b_2 - a_2 b_1)^2.$$

In general, the difference between the two sides is the sum of the squares $(a_i b_j - a_j b_i)^2$ over all pairs $i < j$. A sum of squares is never negative, and it is zero only when every $a_i b_j - a_j b_i$ vanishes, that is, when the lists are proportional. Done.

There is another proof, due to Hermann Schwarz in 1888. The polynomial in $t$

$$p(t) = \sum_i (a_i t + b_i)^2 = \Big(\sum a_i^2\Big) t^2 + 2\Big(\sum a_i b_i\Big) t + \sum b_i^2$$

is never negative, since it is a sum of squares. So its discriminant cannot be positive: $4\left(\sum a_i b_i\right)^2 - 4 \sum a_i^2 \sum b_i^2 \le 0$, which is the inequality.

The means, forwards and backwards

In the same note, Cauchy proved that the geometric mean of positive numbers never exceeds their arithmetic mean:

$$\sqrt[n]{x_1 x_2 \cdots x_n} \le \frac{x_1 + x_2 + \cdots + x_n}{n}.$$

The proof is an induction in two directions. For $n = 2$, just note that

$$\left(\frac{x + y}{2}\right)^2 - xy = \left(\frac{x - y}{2}\right)^2 \ge 0.$$

Forwards: if it holds for $n$ numbers, it holds for $2n$. Split the $2n$ numbers into two halves; each half has product at most $A_1^n$ and $A_2^n$, where $A_1$ and $A_2$ are the means of the halves. By the case $n = 2$, $A_1 A_2 \le \left(\frac{A_1 + A_2}{2}\right)^2$, and $\frac{A_1 + A_2}{2}$ is the mean of all of them. So the inequality holds for 2, 4, 8, 16, …

Backwards: if it holds for $N$ numbers, it holds for any $n < N$. Given $x_1, \ldots, x_n$ with mean $A$, add $N - n$ copies of $A$ itself. The mean of the $N$ numbers is still $A$, and the inequality for $N$ gives

$$x_1 \cdots x_n \cdot A^{N-n} \le A^N, \quad\text{so}\quad x_1 \cdots x_n \le A^n.$$

Since every $n$ lies below some power of 2, the theorem holds for all $n$.

Exercise: the maximum of 3x + 4y

What is the largest value of $3x + 4y$ when $x^2 + y^2 = 1$?

By Cauchy’s inequality, $(3x + 4y)^2 \le (3^2 + 4^2)(x^2 + y^2) = 25$, so $3x + 4y \le 5$. Equality requires $(x, y)$ proportional to $(3, 4)$: $x = \frac{3}{5}$, $y = \frac{4}{5}$, and indeed $3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = \frac{25}{5} = 5$.

Exercise: a product that never drops below 9

Show that $(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9$ for $a, b, c > 0$.

Apply Cauchy’s inequality to the lists $(\sqrt{a}, \sqrt{b}, \sqrt{c})$ and $\left(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}\right)$. The left side becomes $(1 + 1 + 1)^2 = 9$, and the right side is exactly $(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$. Equality holds when $a = b = c$.

Sources: Augustin-Louis Cauchy, Cours d’analyse (1821), note II · J. Michael Steele, The Cauchy–Schwarz Master Class (2004) · Viktor Bunyakovsky, “Sur quelques inégalités concernant les intégrales ordinaires et les intégrales aux différences finies”, Mémoires de l’Académie de Saint-Pétersbourg (1859).

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