Question

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three unit vectors, then prove that: \[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \leq 9 \]

Hint:

This expression achieves its absolute maximum value of 9 when the three vectors point in symmetrically opposing directions in a plane ($120^\circ$ angles from each other), causing their mutual dot products to equal $-\frac{1}{2}$.

Answer 1

Concept: A unit vector has an absolute scalar length or magnitude of exactly 1. Therefore, if $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. We will use the fundamental vector identity relating the magnitude squared of a vector sum or difference to its dot product: \[ |\vec{u}|^2 = \vec{u} \cdot \vec{u} \] Additionally, we rely on the property that the magnitude squared of any vector sum, such as $|\vec{a} + \vec{b} + \vec{c}|^2$, must always be greater than or equal to zero.

Step 1: Expanding the individual squared magnitude terms using dot products.
Let us expand each of the three terms in the given expression using the identity $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$: \[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \] \[ |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c}) \] \[ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) \]

Step 2: Summing the three expanded expressions and substituting the unit values.
Let us add all three equations together: \[ \text{Sum} = \left(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}\right) + \left(|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b}\cdot\vec{c}\right) + \left(|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c}\cdot\vec{a}\right) \] Combine identical magnitude terms: \[ \text{Sum} = 2|\vec{a}|^2 + 2|\vec{b}|^2 + 2|\vec{c}|^2 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \] Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, substitute $|\vec{a}|^2 = 1$, $|\vec{b}|^2 = 1$, and $|\vec{c}|^2 = 1$: \[ \text{Sum} = 2(1) + 2(1) + 2(1) - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \] \[ \text{Sum} = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \quad \cdots (1) \]

Step 3: Using the semi-positive definite property of vector sums to establish an inequality boundary.
Consider the vector sum $(\vec{a} + \vec{b} + \vec{c})$. The square of its magnitude must be non-negative: \[ |\vec{a} + \vec{b} + \vec{c}|^2 \geq 0 \] Expanding this expression gives: \[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0 \] Substitute the unit values ($1+1+1 = 3$): \[ 3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0 \] Rearranging to isolate the dot product sum: \[ 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq -3 \implies -2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \leq 3 \quad \cdots (2) \]

Step 4: Combining the inequality with equation (1).
Now, substitute the maximum upper bound from equation (2) into our original sum from equation (1): \[ \text{Sum} = 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \leq 6 + 3 \] \[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \leq 9 \] This completes the formal mathematical proof. *(Hence Proved)*