Question

Prove that for any two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \), \[ |\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}| \] Also, write the name of this inequality.

Hint:

The Triangle Inequality for vectors states that the magnitude of the sum of two vectors is less than or equal to the sum of their magnitudes.

Answer 1

Step 1: Vector addition and magnitude. 
We are given two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \), and we need to prove the inequality \( |\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}| \). 
This inequality is known as the **Triangle Inequality** for vectors. 
Step 2: Proof of the Triangle Inequality. 
We know that the square of the magnitude of a vector is equal to the dot product of the vector with itself. Therefore, we have: \[ |\mathbf{a} + \mathbf{b}|^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) \] Expanding the dot product: \[ |\mathbf{a} + \mathbf{b}|^2 = \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \] This simplifies to: \[ |\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 \] Step 3: Using the Cauchy-Schwarz inequality. 
From the Cauchy-Schwarz inequality, we know that: \[ \mathbf{a} \cdot \mathbf{b} \leq |\mathbf{a}| |\mathbf{b}| \] Thus: \[ |\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 \leq |\mathbf{a}|^2 + 2 |\mathbf{a}| |\mathbf{b}| + |\mathbf{b}|^2 \] Step 4: Taking square roots. 
Taking the square root of both sides, we get: \[ |\mathbf{a} + \mathbf{b}| \leq \sqrt{|\mathbf{a}|^2 + 2 |\mathbf{a}| |\mathbf{b}| + |\mathbf{b}|^2} \] Since \( \sqrt{|\mathbf{a}|^2 + 2 |\mathbf{a}| |\mathbf{b}| + |\mathbf{b}|^2} = |\mathbf{a}| + |\mathbf{b}| \), we have: \[ |\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}| \] Step 5: Conclusion. 
Thus, the inequality is proven, and it is known as the **Triangle Inequality** for vectors.