Q.Establish the following vector inequalities geometrically or otherwise:

Concept: Triangle inequality (vector form). Placing $\vec{a}$ and $\vec{b}$ tail-to-head, $\vec{a}+\vec{b}$ is the third side of a triangle whose other sides have lengths $|\vec{a}|$ and $|\vec{b}|$ — a triangle's side is always between the sum and the (positive) difference of the other two.

1. (a) and (b) follow directly from this triangle property applied to $\vec{a}+\vec{b}$.
2. (c) and (d) follow by replacing $\vec{b}$ with $-\vec{b}$ in (a) and (b), since $|-\vec{b}| = |\vec{b}|$.

Equality conditions (verified via $|\vec{a}\pm\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2\pm2|\vec{a}||\vec{b}|\cos\theta$): …

The core idea

Placing $\vec{a}$ and $\vec{b}$ tail-to-head, their sum $\vec{a}+\vec{b}$ is the third side of a triangle whose other two sides have lengths $|\vec{a}|$ and $|\vec{b}|$. In any triangle, one side is at most the sum, and at least the (positive) difference, of the other two sides — that single geometric fact proves all four inequalities.

(a) $|\vec{a}+\vec{b}| \le |\vec{a}|+|\vec{b}|$

Geometrically: the direct side of the triangle ($|\vec{a}+\vec{b}|$) cannot exceed the sum of the other two sides.

Algebraically:

Since $\cos\theta \le 1$:

Taking square roots gives the inequality.

Equality requires $\cos\theta=1$, i.e. $\theta=0^\circ$: $\vec{a}$ and $\vec{b}$ point in the same direction.

(b) $|\vec{a}+\vec{b}| \ge \big||\vec{a}|-|\vec{b}|\big|$

Apply (a) with $\vec{a}$ replaced by $(\vec{a}+\vec{b})+(-\vec{b})$:

Swapping the roles of $\vec{a}$ and $\vec{b}$ similarly gives $|\vec{a}+\vec{b}| \ge |\vec{b}|-|\vec{a}|$. Since $|\vec{a}+\vec{b}|$ is at least both of these values, it must be at least their absolute value:

Equality requires $\vec{a}$ and $\vec{b}$ to point in opposite directions ($\theta=180^\circ$), where the resultant's magnitude is exactly the difference of the two magnitudes.

(c) $|\vec{a}-\vec{b}| \le |\vec{a}|+|\vec{b}|$

Replace $\vec{b}$ with $-\vec{b}$ in (a). Since $|-\vec{b}|=|\vec{b}|$:

Equality requires $\vec{a}$ and $-\vec{b}$ to point in the same direction, i.e. $\vec{a}$ and $\vec{b}$ point in opposite directions.

(d) $|\vec{a}-\vec{b}| \ge \big||\vec{a}|-|\vec{b}|\big|$

Replace $\vec{b}$ with $-\vec{b}$ in (b): …

Concept: The Triangle Inequality Theorem of Euclidean Geometry

Method: Pure Geometric Proof (No Dot Products, No Components) -- the "Otherwise" Route

This method proves all four inequalities using only a classical fact from Euclidean geometry -- in any triangle, one side is at most the sum, and at least the (positive) difference, of the other two sides, with equality exactly when the triangle degenerates into a straight line -- applied to the triangle formed by placing the vectors head-to-tail. No dot products, no components, no trigonometric identities are used anywhere.

Setting up the triangle

Place $\vec a$ with its tail at point $O$ and head at point $P$. Place $\vec b$ with its tail at $P$ and head at point $Q$. Then $\vec a+\vec b$ is exactly the vector $\overrightarrow{OQ}$ -- the third side of triangle $OPQ$, whose other two sides are $\overrightarrow{OP}=\vec a$ (length $|\vec a|$) and $\overrightarrow{PQ}=\vec b$ (length $|\vec b|$).

(a) $|\vec a+\vec b| \le |\vec a|+|\vec b|$

By the Triangle Inequality Theorem, side $OQ$ of the triangle can never exceed the sum of the other two sides:

Equality occurs exactly when the "triangle" degenerates into a straight line with $P$ lying between $O$ and $Q$ -- geometrically, this happens precisely when $\vec a$ and $\vec b$ point in the same direction.

(b) $|\vec a+\vec b| \ge \big||\vec a|-|\vec b|\big|$

The same Triangle Inequality Theorem also states that any side is at least the difference of the other two: $OQ \ge |OP-PQ|$ (otherwise the two shorter sides couldn't reach around to close the triangle at all). So:

Equality occurs when the triangle again degenerates to a straight line, but this time with $O$ and $Q$ on the same side of $P$ (i.e. $P$ is not between them) -- geometrically, $\vec a$ and $\vec b$ point in opposite directions.

(c) and (d) -- replace $\vec b$ with the vector $-\vec b$

Construct $-\vec b$: same length as $\vec b$ ($|-\vec b|=|\vec b|$), pointing the opposite way. Since $\vec a - \vec b = \vec a + (-\vec b)$, this is literally the same geometric picture as (a) and (b), with the second side of the triangle relabelled from $\vec b$ to $-\vec b$:

$|\vec a-\vec b| = |\vec a+(-\vec b)| \ge \big||\vec a|-|-\vec b|\big| = \big||\vec a|-|\vec b|\big| \qquad \text{(d)}$ …