triangular inequality

Question

If we write $||a+b||\leq||a||+||b||$ explicitly in $\mathbb{R}^n$ it is $\sqrt{\sum^n_1(a_i+b_i)^2}\leq \sqrt{\sum^n_1(a_i)^2}+\sqrt{\sum^n_1(b_i)^2}$ how can it be if $\sqrt{(a_i)^2}+\sqrt{(b_i)^2}\leq\sqrt{(a_i+b_i)^2}$ (for every $i$)?

Answer 1

As it happens, the identity you give: $$\sqrt{a_i^2}+\sqrt{b_i^2}\le\sqrt{(a_i+b_i)^2}$$ is false. Take for example $a_i=1$ and $b_i=-1$ to see why it is false.