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Prove Minkowski's inequality using Cauchy-Schwarz's

  • #1
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Homework Statement


For u and v in [itex]R^n[/itex] prove Minkowski's inequality that [itex]\|u + v\| \leq \|u\| + \|v\|[/itex] using the Cauchy-Schwarz inequality theorem: [itex]|u \cdot v| \leq \|u\| \|v\|[/itex].

Homework Equations


Dot product: [itex]u \cdot v = u_1 v_1 + u_2 v_2 + \cdots + u_n v_n[/itex]
Norm: [itex]\|u \| = \sqrt {u \cdot u}[/itex]
Cauchy-Schwarz inequality: [itex]|u \cdot v| \leq \|u\| \|v\|[/itex]

The Attempt at a Solution


Using def. of norm: [itex]\|u + v\|^2 = \sqrt {(u + v) \cdot (u + v)}^2 = (u + v) \cdot (u + v)[/itex]
Expand: [itex](u + v) \cdot (u + v) = u \cdot u + 2 (u \cdot v) + v \cdot v[/itex]
Using the Cauchy-Schwarz inequality: [itex]u \cdot u + 2 (u \cdot v) + v \cdot v \leq \|u \|^2 + 2 \|u \| \|v\| + \| v \|^2 = (\|u \| + \| v \|)^2[/itex]

Therefore, [itex]\|u + v\|^2 \leq (\|u \| + \| v \|)^2 \Rightarrow \|u + v\| \leq \|u \| + \| v \|[/itex].

Thank-you
 

Answers and Replies

  • #2
jbunniii
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Looks fine to me.
 

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