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Prove that if $n≥1$ and $a_1,a_2,….,a_n$ are any real numbers, then $|a_1+a_2+⋯+a_n |≤|a_1 |+|a_2 |+⋯+|a_n |$.
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The usual induction approach, but you need something special at $n=2$. It is clearly true for $n=1$. Now prove $n=2$ from the triangle inequality. Then if it is true for $k$ numbers, prove it is true for $k+1$ numbers by grouping the first $k$ into one group and combine them with the last one using your proof of $n=2$
Ross Millikan
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The base case is trivial. Now suppose we have already proven the assertion for $n\ge 1$. So $|\sum_{i=1}^{n+1}a_i|=|a_{n+1}+\sum_{i=1}^{n}a_i|\le |a_{n+1}|+|\sum_{i=1}^{n}a_i|\le|a_{n+1}|+\sum_{i=1}^{n}|a_i|=\sum_{i=1}^{n+1}|a_i|$, which closed the induction.
(Note that we use our inductive hypothesis $|\sum_{i=1}^{n}a_i|\le \sum_{i=1}^{n}|a_i|$ above)
So you only need to show that $|a+b|\le |a|+|b|$
Jose Antonio
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