Q.For any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|$ (triangle inequality).

Concept: Triangle Inequality — the magnitude of a sum is at most the sum of the magnitudes.

Reasoning:

1. Start from the squared magnitude: $|\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$.
2. By the Cauchy–Schwarz inequality, $\vec{a}\cdot\vec{b} \le |\vec{a}||\vec{b}|$, so $|\vec{a}+\vec{b}|^2 \le |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}| = (|\vec{a}|+|\vec{b}|)^2$. …

The triangle inequality is one of the most intuitive yet powerful results in vector algebra. It simply says: the length of one side of a triangle cannot exceed the sum of the lengths of the other two sides. When you add two vectors $\vec{a}$ and $\vec{b}$, the resultant $\vec{a}+\vec{b}$ forms the third side of a triangle whose other two sides are $\vec{a}$ and $\vec{b}$.

So the statement $|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|$ is not just a formula — it’s a geometric fact. It holds for any two vectors, regardless of direction. There is no exception.

Let’s see why this is always true.

1. Start with the definition of magnitude squared. For any vectors $\vec{a}$ and $\vec{b}$,

2. Use the Cauchy-Schwarz inequality. The dot product satisfies $\vec{a}\cdot\vec{b} \le |\vec{a}||\vec{b}|$. So

3. Take the square root (both sides are non-negative). …

Method: Proving a magnitude inequality by squaring

For any inequality about the length of a vector sum, magnitudes are awkward to manipulate directly but their squares expand cleanly through the dot product. Square first, bound the cross term, then take the root.

Steps

Step 1: Square the magnitude and expand via the dot product

Squaring is the key: $|\ \cdot\ |^2$ becomes a dot product, which distributes just like ordinary multiplication.

Step 2: Bound the one term you don't control

The only unknown term is $\vec{a}\cdot\vec{b}$. Bound it above with Cauchy–Schwarz, $\vec{a}\cdot\vec{b} \le |\vec{a}\cdot\vec{b}| \le |\vec{a}||\vec{b}|$: …

Common Mistakes

Mistake 1: Treating magnitude as if it were linear, i.e. $|\vec{a}+\vec{b}| = |\vec{a}|+|\vec{b}|$.

Why it's wrong: this equality holds only when the vectors point in the same direction; in general the left side is strictly smaller. Correct approach: it is an inequality ($\le$), and equality is a special case, not the rule.

Mistake 2: Forgetting the cross term when squaring.

Why it's wrong: $(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})$ expands to $|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}$ — dropping the $2\vec{a}\cdot\vec{b}$ term breaks the whole proof. Correct approach: expand the dot product fully, then bound only that middle term. …