Q.Show that $|\vec{a} + \vec{b}| < |\vec{a}| + |\vec{b}|$ is always true for any two vectors $\vec{a}$ and $\vec{b}$.

The Triangle Inequality

The shortest way from one point to another is a straight line. Any detour through a third point can only make the trip longer — never shorter. That everyday fact is the triangle inequality.

The Intuition

Stretch a string between two points and it snaps straight — the shortest possible path. Introduce a bend, or route the string through some middle point $B$, and the total length grows. So going from $A$ to $C$ directly is never longer than going $A \to B \to C$.

The Statement

For any three points $A$, $B$, $C$:

Equality holds only when $B$ lies on the segment $AC$ — the detour is along the same straight line, so nothing is wasted. Applied to a genuine triangle (three non-collinear points), each side is strictly less than the sum of the other two, which is why sides $3, 4, 10$ cannot form a triangle: $10 > 3 + 4$.

The Vector Form

Since a displacement from $A$ to $C$ can be split as $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$, the inequality becomes a statement about vector lengths:

The length of a sum of two vectors is at most the sum of their lengths, with equality only when $\vec{a}$ and $\vec{b}$ point in the same direction. There is also a companion lower bound, $|\vec{a}+\vec{b}| \ge \big||\vec{a}|-|\vec{b}|\big|$, which comes from the same idea applied to $\vec{a}=(\vec{a}+\vec{b})+(-\vec{b})$. …