I am abjectly disappointed that I could not prove this statement on my own. I have tried it directly and by contradiction but hit a wall. Here is the statement (again) and my proof (thus far):
Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.
Proof: Let $a_{1}, ..., a_{n}$ be real numbers. Consider $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$. Then $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}| \leq ||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}||$ holds by a property of the Absolute Value function $\forall a_{1},...,a_{n} \in \mathbb{R}$. Now, by the Reverse Triangle Inequality, we have $||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|| \leq |a_{1} - \sum\limits_{k=2}^{n} a_{k}|$.
I then proceeded to show that $|a_{1} - \sum\limits_{k=2}^{n} a_{k}| \leq \sum\limits_{k=1}^{n} a_{k} \leq |\sum\limits_{k=1}^{n} a_{k}|$ by considering different cases for $a_{1}$ and $\sum\limits_{k=1}^{n} a_{k}$ (e.g. $a_{1}, \sum\limits_{k=1}^{n} a_{k} > 0$ as one case out of four), and was able to show one case to hold but am struggling with the others. Is there a more simplified way to write this proof?
Any help would be greatly appreciated! This has hurt my brain for weeks now :(
We have $$\left|\sum_{k = 1}^n a_k\right| = \left|a_1 + \sum_{k = 2}^n a_k\right| \ge |a_1| - \left|\sum_{k = 2}^n a_k\right| \ge |a_1| - \sum_{k= 2}^n|a_k|,$$ where the first inequality comes from $|a+b| \ge |a| - |b|$ and the second one comes from $|a + b| \le |a| + |b|$.
EDIT: To prove that $|a+b| \ge |a| - |b|$ just remark that $$|a| = |a + b - b| \le |a+b| + |b|$$ by triangular inequality.