Triangle inequality

Submitted by Structure on Sat, 12/04/2010 - 08:54.

Let us show that for

$x_i, y_i \in R$

we have

$\sqrt{\sum_{i=1 }^ n(x_i+y_i)^2}\le\sqrt{\sum_{i=1 }^n x_i^2}+\sqrt{\sum_{i=1 }^ny_i^2}$

We have

$(x_i+y_i)^2=x_i^2+2x_iy_i+y_i^2$

so we also have

$\sum_{i=1 }^ n(x_i+y_i)^2=\sum_{i=1 }^ nx_i^2+2\sum_{i=1 }^ nx_iy_i+\sum_{i=1 }^ ny_i^2\le\sum_{i=1 }^ nx_i^2+2\sqrt{\sum_{i=1 }^n x_i^2}\sqrt{\sum_{i=1 }^ny_i^2}+\sum_{i=1 }^ ny_i^2=$

$=\left(\sqrt{\sum_{i=1 }^n x_i^2}+\sqrt{\sum_{i=1 }^ny_i^2}\right )^2$

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Triangle inequality