Classic Inequalities: Cauchy-Schwarz and AM-GM

Some questions ask for the exact answer. Others only ask how big can this possibly get? — how large a correlation is even legal, how fast a portfolio can compound, whether a covariance matrix someone handed you is physically possible. Those do not need solving; they need a ceiling. Two inequalities supply almost every ceiling a quant needs, and they turn up constantly in interviews because they answer in one line what looks like a page of calculus.

First, a picture with no symbols in it

Stand a pencil upright in sunlight and look at its shadow on the desk. The shadow can be as long as the pencil, but never longer — and it reaches full length only when the pencil is lying flat, pointing exactly along the shadow's direction. Tilt the pencil at all and the shadow shortens. That is Cauchy-Schwarz: a quantity built from two things is capped by the sizes of those two things, and the cap is reached only when the two point in exactly the same direction.

Now the second picture. You have 40 metres of fencing and want to enclose the biggest rectangular paddock you can. A long thin strip, 19 m by 1 m, encloses 19 m². A square, 10 m by 10 m, encloses 100 m². Same fence, five times the land. That is AM-GM: for a fixed sum, the product is largest when the pieces are all equal, and lopsidedness always costs you.

Hold both pictures. Every formula below is one of them written down.

Cauchy-Schwarz, written out

Take two lists of numbers of the same length. Call them $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ — think of them as two vectors, or two columns in a spreadsheet. Then

$$\left( \sum_{i=1}^{n} a_i b_i \right)^{\!2} \;\le\; \left( \sum_{i=1}^{n} a_i^2 \right) \left( \sum_{i=1}^{n} b_i^2 \right).$$

Read out loud: the square of "multiply the lists together term by term and add" is never bigger than "the sum of squares of the first list" times "the sum of squares of the second list". The symbol $\sum_{i=1}^n$ just means "add up over every position $i$ from 1 to $n$". Equality happens in exactly one case: when one list is a constant multiple of the other, $b_i = \lambda a_i$ for every $i$ — the pencil lying flat.

Worked example 1: the two lists, by hand

Let $a = (3, 4)$ and $b = (1, 2)$.

• Left side. Term by term: $3 \times 1 = 3$ and $4 \times 2 = 8$. Sum: $3 + 8 = 11$. Square it: $11^2 = 121$.
• Right side. Sum of squares of $a$: $9 + 16 = 25$. Sum of squares of $b$: $1 + 4 = 5$. Product: $25 \times 5 = 125$.
• Check: $121 \le 125$. True, with a small gap of 4.

Now make the lists proportional. Keep $a = (3,4)$ and set $b = (6, 8)$, which is exactly $2a$.

• Left side: $3 \times 6 = 18$, $4 \times 8 = 32$, sum $= 50$, squared $= 2500$.
• Right side: $25 \times (36 + 64) = 25 \times 100 = 2500$.
• Check: $2500 \le 2500$. Equality, because the lists point the same way.

The gap in the first case measures how much the two lists disagree in direction. That is not an analogy — it is the definition of correlation, which is the next section.

The same inequality in probability language

Swap "add over positions" for "average over outcomes" and nothing else changes. For two random quantities $X$ and $Y$,

$$\big( E[XY] \big)^2 \;\le\; E[X^2]\, E[Y^2],$$

where $E[\cdot]$ is the expected value, the probability-weighted average. In words: the average of the product is capped by the sizes (in a root-mean-square sense) of the two things being multiplied. Apply this to deviations from the mean and the left side becomes the covariance while the right becomes the product of the two variances:

$$\operatorname{Cov}(X,Y)^2 \;\le\; \operatorname{Var}(X)\, \operatorname{Var}(Y).$$

Divide both sides by the right-hand side and take a square root, and you get $|\rho| \le 1$: the correlation coefficient can never leave the range $-1$ to $+1$. That famous fact is not a separate rule. It is Cauchy-Schwarz.

Drag the correlation slider below. The scatter never becomes a perfect line until $\rho$ reaches exactly $\pm 1$ — the equality case, where one variable is a fixed multiple of the other.

Worked example 2: is this covariance matrix even possible?

A risk system reports, for two strategies: daily volatility $\sigma_A = 6\%$ and $\sigma_B = 5\%$, with covariance $\operatorname{Cov}(A,B) = 0.0040$.

• Convert the volatilities to variances: $\operatorname{Var}(A) = 0.06^2 = 0.0036$ and $\operatorname{Var}(B) = 0.05^2 = 0.0025$.
• Cauchy-Schwarz ceiling on the covariance: $\sqrt{0.0036 \times 0.0025} = \sqrt{0.000009} = 0.0030$.
• Reported covariance is $0.0040$, which exceeds $0.0030$.

So the numbers are impossible — they imply $\rho = 0.0040 / 0.0030 = 1.33$. Something in the pipeline is broken: different sample windows, a stale volatility estimate, or a hand-edited cell. This check takes ten seconds and catches real bugs; it is why nearest-correlation-matrix repair exists and why positive semidefinite is the condition a covariance matrix must satisfy.

AM-GM, written out

For any non-negative numbers $x_1, x_2, \dots, x_n$, the arithmetic mean (add them up, divide by how many) is never below the geometric mean (multiply them together, take the $n$-th root):

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

In words: averaging by adding always gives you at least as much as averaging by multiplying, and the two are equal only when all the numbers are identical. The more spread out the numbers, the wider the gap. Here is the classic picture, a semicircle: the radius is the arithmetic mean of $a$ and $b$, the vertical bar is their geometric mean, and a bar inside a circle simply cannot be taller than the radius.

Worked example 3: why +50% then -50% loses money

A fund gains 50% one year and loses 50% the next. The two growth factors are $1.5$ and $0.5$.

• Arithmetic mean of the factors: $(1.5 + 0.5)/2 = 1.00$, which looks like "flat".
• Geometric mean: $\sqrt{1.5 \times 0.5} = \sqrt{0.75} = 0.8660$.
• AM-GM check: $1.00 \ge 0.8660$. The gap is $0.134$, and that gap is your loss.

Follow the actual money. Start with $100. After year one: $100 \times 1.5 = 150$. After year two: $150 \times 0.5 = 75$. You are down 25%. Cross-check with the geometric mean: $100 \times 0.8660^2 = 100 \times 0.75 = 75$. It matches, because the geometric mean is by construction the constant rate that reproduces the final wealth.

This is volatility drag, and AM-GM says it is unavoidable: the compounded return is always at or below the average return, and the shortfall grows with dispersion. It is why Kelly sizing exists, and it is the same curvature effect as Jensen's inequality.

Worked example 4: optimisation with no derivatives

Minimise $f(x) = x + \dfrac{4}{x}$ for $x > 0$. Apply AM-GM to the two terms $x$ and $4/x$:

$$\frac{x + 4/x}{2} \;\ge\; \sqrt{x \cdot \frac{4}{x}} = \sqrt{4} = 2 .$$

The product inside the root is $4$ whatever $x$ is — the $x$ cancels — so the right-hand side is a fixed number. Multiply through by 2: $x + 4/x \ge 4$. Equality needs the two terms equal: $x = 4/x$, so $x^2 = 4$, so $x = 2$.

Sanity-check by hand: $x = 1$ gives $1 + 4 = 5$; $x = 2$ gives $2 + 2 = 4$; $x = 4$ gives $4 + 1 = 5$. The minimum is 4 at $x = 2$, found without differentiating anything.

What this means in practice

• Data validation. Any reported covariance above $\sigma_A \sigma_B$ is a bug, not a signal. Same for a correlation outside $[-1, 1]$ or a risk contribution larger than total risk.
• Compounding. Report geometric returns, not arithmetic ones, when describing what an investor actually earned. The gap between the two measures how bumpy the ride was.
• Sizing and diversification. AM-GM is the formal reason equal-ish weights across similar assets beat concentrated ones for a fixed risk budget; Cauchy-Schwarz bounds how much diversification a correlated basket can give you.
• Interviews. "Maximise $xy$ subject to $x + y = 10$" and "how large can $\sum a_i b_i$ be?" are these two in costume. Answer in one line, then state the equality case — that is what the interviewer is listening for.