Exercise 1025Linear algebra › Inner product spacesHigh schoolAdded 3 Nov 2026

Triangle inequality

#triangle inequality#norm#inner product

Statement

Let $\|\cdot\|$ be the norm associated with an inner product. Show that
$\|u+v\|\le\|u\|+\|v\|.$

Hint

Expand $\|u+v\|^2$ and bound $\langle u,v\rangle$ using Cauchy-Schwarz.

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Solution

Expansion.
$\|u+v\|^2=\|u\|^2+2\langle u,v\rangle+\|v\|^2.$

Bound. By Cauchy-Schwarz, $\langle u,v\rangle\le|\langle u,v\rangle|\le\|u\|\|v\|$ , so
$\|u+v\|^2\le\|u\|^2+2\|u\|\|v\|+\|v\|^2=(\|u\|+\|v\|)^2.$

Conclusion. Taking square roots: $\|u+v\|\le\|u\|+\|v\|$ , with equality iff $u,v$ are nonnegatively proportional.

Triangle inequality

Statement

Solution

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