Triangle inequality

Dependencies:

1. Inner product space
2. Cauchy-Schwarz Inequality
3. /complex-numbers/real-part-less-than-abs

Let $V$ be an inner product space over $F$.

Then $\|x+y\| \le \|x\| + \|y\|$.

Proof

\begin{align} & (\|x\| + \|y\|)^2 - \|x + y\|^2 \\ &= (\|x\|^2 + \|y\|^2 + 2\|x\|\|y\|) - (\|x\|^2 + \|y\|^2 + \langle x, y \rangle + \langle y, x \rangle) \\ &= 2(\|x\|\|y\| - (\langle x, y \rangle + \overline{\langle x, y \rangle})/2) \\ &= 2(\|x\|\|y\| - \operatorname{Re}(\langle x, y \rangle)) \\ &\ge 2(\|x\|\|y\| - |\langle x, y \rangle|) \\ &\ge 0 \tag{by Cauchy-Schwarz inequality} \end{align}

Therefore, $\|x+y\|^2 \le (\|x\| + \|y\|)^2$. Since both $\|x+y\|$ and $\|x\| + \|y\|$ are non-negative, $\|x+y\| \le \|x\| + \|y\|$.

Dependency for: None

Info:

• Depth: 7
• Number of transitive dependencies: 10

Transitive dependencies:

1. /complex-numbers/conjugation-is-homomorphic
2. /complex-numbers/real-part-less-than-abs
3. Group
4. Ring
5. Field
6. Vector Space
7. Inner product space
8. Inner product is anti-linear in second argument
9. Zero in inner product
10. Cauchy-Schwarz Inequality