Q.For any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a}\cdot\vec{b}|\le|\vec{a}|\,|\vec{b}|$ (Cauchy-Schwartz inequality).

Claim (Cauchy–Schwarz): for any two vectors, $|\vec{a}\cdot\vec{b}| \le |\vec{a}|\,|\vec{b}|$.

Step 1 — Use the angle form of the dot product. For any $\vec{a}$, $\vec{b}$ with angle $\theta$ between them,

Step 2 — Take magnitudes.

Step 3 — Bound the cosine. Since $|\cos\theta| \le 1$ always,

$|\vec{a}\cdot\vec{b}| = |\vec{a}|\,|\vec{b}|\,|\cos\theta| \le |\vec{a}|\,|\vec{b}|.$ …

The Cauchy–Schwarz inequality bounds how large a dot product can be: it can never exceed the product of the two lengths. The dot product measures alignment, and the most two vectors can align is to point exactly the same way — that limiting case is what makes the inequality tight.

1. The two faces of the dot product

The dot product has a component form and a geometric form, and both give the same number:

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. The geometric form is the one we need, because it shows the dot product as a length product scaled by a cosine.

2. Take magnitudes

Taking the absolute value of both sides and using that lengths are non-negative,

3. The cosine is at most 1

For any real angle $\theta$, $-1 \le \cos\theta \le 1$, so $|\cos\theta| \le 1$. Multiplying the non-negative quantity $|\vec{a}||\vec{b}|$ by something at most $1$ can only shrink it:

Chaining the last two lines,

which is exactly the Cauchy–Schwarz inequality. …

Method: Proving a dot-product bound via the angle form

Use this whenever you must prove an inequality that bounds a dot product: the trick is to switch from the component form to the geometric form, so that a bounded trigonometric ratio does all the work.

Steps

Step 1: Replace the dot product by its geometric form

Any dot product can be written as $\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$, where $\theta$ is the angle between the vectors. This move is the whole idea: the lengths are fixed non-negative numbers, so all the variation is carried by $\cos\theta$.

Step 2: Take magnitudes and isolate the trig factor

Step 3: Apply the standard bound on cosine …

Common Mistakes

Mistake 1: Dropping the absolute value and writing $\vec{a}\cdot\vec{b} \le |\vec{a}||\vec{b}|$ as the claim.

Why it's wrong: a dot product can be negative, so bounding only $\vec{a}\cdot\vec{b}$ (not $|\vec{a}\cdot\vec{b}|$) misses half the statement. Correct approach: the inequality is about the magnitude $|\vec{a}\cdot\vec{b}|$, which is why $|\cos\theta| \le 1$ (with the bars) is the step that matters.

Mistake 2: Confusing this with the triangle inequality. …