Triangle inequality

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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