I want to show that $\forall a,b\in\mathbb{R}$, we have $|a|+|b|\ge|a+b|$, using the piecewise definition of the absolute value. In my attempt, I have broken the problem up into many cases. All of this casework has me wondering if I've made some subtle mistake somewhere, especially with the last two cases, and so I would appreciate it if someone would look over the proof and see whether or not it is acceptable. Have I accounted for every possible case? Is there a flaw in my reasoning somewhere, especially with cases $3$ & $4$? Thanks in advance.
My proof:
Case 1, $a,b\ge0$:
By definition, $|a|+|b|=a+b$.
Since $a,b\ge0$, we have that $a+b\ge0$, and therefore, by definition, $|a+b|=a+b$.
Thus we have $|a|+|b|=|a+b|$, which implies that $|a|+|b|\ge|a+b|$.
So, $\forall a,b\in\mathbb{R}$ such that $a,b\ge0$, we have $|a|+|b|\ge|a+b|$. $\blacksquare$
Case 2, $a,b<0$:
By definition, $|a|+|b|=(-a)+(-b)=-a-b$.
Since $a,b<0$, we have that $a+b<0$, and therefore, by definition, $|a+b|=-(a+b)=-a-b$.
Thus we have $|a|+|b|=|a+b|$, which implies that $|a|+|b|\ge|a+b|$.
So, $\forall a,b\in\mathbb{R}$ such that $a,b<0$, we have $|a|+|b|\ge|a+b|$. $\blacksquare$
Case 3, $a\ge0,b<0$:
By definition, $|a|+|b|=a-b$.
If $|a|\ge|b|$, then $a+b$ is non-negative, and thus, by definition, $|a+b|=a+b$. Since $b$ is always strictly negative, $-b>b$, and therefore $a-b>a+b$, which implies that $|a|+|b|>|a+b|$ and that $|a|+|b|\ge|a+b|$.
If $|a|\le|b|$, then $a+b$ is non-positive, and thus, by definition, $|a+b|=-(a+b)=-a-b$. Since $a$ is always non-negative, $a>-a$, and therefore $a-b>-a-b$, which implies that $|a|+|b|>|a+b|$ and that $|a|+|b|\ge |a+b|$.
So, $\forall a,b\in\mathbb{R}$ such that $a\ge0, b<0$, we have $|a|+|b|\ge|a+b|$. $\blacksquare$
Case 4, $a<0,b\ge0$:
By definition, $|a|+|b|=-a+b$.
If $|a|\ge|b|$, then $a+b$ is non-positive, and thus, by definition, $|a+b|=-(a+b)=-a-b$. Since $b$ is always non-negative, $b\ge-b$, and therefore $-a+b\ge-a-b$, which implies that $|a|+|b|\ge|a+b|$.
If $|a|\le|b|$, then $a+b$ is non-negative, and thus, by definition, $|a+b|=a+b$. Since $a$ is strictly negative, $-a>a$, and therefore, $-a+b>a+b$, which implies that $|a|+|b|>|a+b|$ and that $|a|+|b|\ge|a+b|$.
So, $\forall a,b\in\mathbb{R}$ such that $a<0,b\ge0$, we have $|a|+|b|\ge|a+b|$. $\blacksquare$
And this result completes the proof for the general case. $\square$