Proving One Sum is Greater Than Another Ok so I'm looking to prove that $$\left|\sum_{j=1}^n a_j\right|\geq|a_1|-\sum_{j=2}^n|a_j|.$$ My first instinct is to try to use the reverse triangle inequality but it does not seem to be working. I know that $$\left|\sum_{j=1}^n a_j\right|\leq\sum_{j=1}^n|a_j|.$$ I just can't see where that fact comes into play. Any help is appreciated thank you.
 A: Detailed (from the basic triangle inequality)
From the triangle inequality:
$$
\lvert a+ b \rvert \leq \lvert a \rvert+ \lvert b \rvert \tag{1}
$$
you get
$$
\lvert a \rvert = \lvert a+ b - b \rvert \leq \lvert a +b \rvert+ \lvert -b \rvert=\lvert a +b \rvert+ \lvert b \rvert
$$
i.e. the "reverse triangle inequality"
$$
\lvert a \rvert - \lvert b \rvert \leq \lvert a +b \rvert \tag{2}
$$
Now, take $a\stackrel{\rm def}{=} a_1$, $b\stackrel{\rm def}{=} \sum_{j=2}^n a_j$ to obtain
$$
\lvert a_1 \rvert - \left\lvert \sum_{j=2}^n a_j \right\rvert \leq \left\lvert \sum_{j=1}^n a_j \right\rvert \tag{3}
$$
Since by the "regular" triangle inequality, $\left\lvert \sum_{j=1}^n a_j \right\rvert \leq \sum_{j=1}^n \left\lvert a_j \right\rvert$, we finally have:
$$
\lvert a_1 \rvert - \sum_{j=2}^n \left\lvert a_j \right\rvert \leq
\lvert a_1 \rvert - \left\lvert \sum_{j=2}^n a_j \right\rvert \leq \left\lvert \sum_{j=1}^n a_j \right\rvert \tag{4}
$$
A: This is just a combination of the reverse triangle inequality and the usual triangle inequality.  To keep things straight, let $b=\sum_{j=1}^n a_j$ and $c=\sum_{j=2}^n a_j$.  Then $b=a_1+c$, so by the reverse triangle inequality, $|b|\geq |a_1|-|c|$.  Now you want to use the triangle inequality on the $|c|$ term to finish; I'll let you work out the details.
