Prove That for Any Two Vectors a and b, |a + b| ≤ |a| + |b|

A standard Class 12 vector algebra proof, not tied to a specific exam paper.

Problem

Prove that for any two vectors $\vec a$ and $\vec b$: $|\vec a+\vec b| \leq |\vec a|+|\vec b|$.

Key insight. Expanding $|\vec a+\vec b|^2$ via the dot product brings in the cross term $2\,\vec a\cdot\vec b$ — and since a dot product can never exceed the product of the two vectors’ magnitudes, that single bound is enough to compare $|\vec a+\vec b|^2$ against $(|\vec a|+|\vec b|)^2$ directly.

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Approach

Expand $|\vec a+\vec b|^2$ using the dot product $(\vec a+\vec b)\cdot(\vec a+\vec b)$. Bound the resulting cross term $\vec a\cdot\vec b$ using the fact that $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ can never exceed $|\vec a||\vec b|$. This shows $|\vec a+\vec b|^2\leq(|\vec a|+|\vec b|)^2$, and taking square roots (both sides being non-negative) completes the proof.

Solution

Step 1 — Expand |a + b|² using the dot product

$$|\vec a+\vec b|^2 = (\vec a+\vec b)\cdot(\vec a+\vec b) = \vec a\cdot\vec a + 2\,\vec a\cdot\vec b + \vec b\cdot\vec b = |\vec a|^2+2\,\vec a\cdot\vec b+|\vec b|^2$$

Step 2 — Bound the dot product term

Since $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$, where $\theta$ is the angle between $\vec a$ and $\vec b$, and $\cos\theta\leq1$ always:

$$\vec a\cdot\vec b \leq |\vec a||\vec b|$$

Step 3 — Apply this bound to the expansion

$$|\vec a+\vec b|^2 \leq |\vec a|^2+2|\vec a||\vec b|+|\vec b|^2$$

Step 4 — Recognise the right-hand side as a perfect square

$$|\vec a|^2+2|\vec a||\vec b|+|\vec b|^2 = \left(|\vec a|+|\vec b|\right)^2$$

Step 5 — Combine and take square roots

$$|\vec a+\vec b|^2 \leq \left(|\vec a|+|\vec b|\right)^2$$

Since both $|\vec a+\vec b|$ and $|\vec a|+|\vec b|$ are non-negative, taking the square root of both sides preserves the inequality:

$$|\vec a+\vec b| \leq |\vec a|+|\vec b|$$

$\blacksquare$

Answer

Common mistakes

• Trying to prove the inequality by squaring $|\vec a|+|\vec b|$ and $|\vec a+\vec b|$ separately without connecting them through the dot-product expansion. The whole proof hinges on recognising that $|\vec a+\vec b|^2$ naturally expands to include the same $|\vec a|^2$, $|\vec b|^2$, and a cross term — the cross term is exactly what needs bounding, and it’s only visible after this expansion.
• Forgetting that $\cos\theta$ can be negative. The bound $\vec a\cdot\vec b\leq|\vec a||\vec b|$ still holds even when $\cos\theta<0$ (making $\vec a\cdot\vec b$ negative), since a negative number is automatically less than the positive quantity $|\vec a||\vec b|$ — the inequality doesn’t require $\theta$ to be acute.

Practise next

• Using the same dot-product expansion technique, prove the reverse triangle inequality: $\big||\vec a|-|\vec b|\big| \leq |\vec a-\vec b|$.