Consequence of triangular inequality

Let $p: \mathbb{R}^{2} \rightarrow \mathbb{R}$ be defined by $$ p(x, y)=\left\{\begin{array}{lll} |x| & \text { if } & x \neq 0 \\ |y| & \text { if } & x=0 . \end{array}\right. $$ Which of the following statements are true?

1. $p(x, y)=0$ if and only if $x=y=0$
2. $p(x, y) \geq 0$ for all $x, y$
3. $p(\alpha x, \alpha y)=|\alpha| p(x, y)$ for all $\alpha \in \mathbb{R}$ and for all $x, y$
4. $p\left(x_{1}+x_{2}, y_{1}+y_{2}\right) \leq p\left(x_{1}, y_{1}\right)+p\left(x_{2}, y_{2}\right)$ for all $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$

Options $1$, $2$ and $3$ are true immediately from the definition of $p$ itself, but, is there any counter for $4$? I got $4$ is true when $x_1=x_2=0$ by the triangular inequality. But what about other cases?

Solution 1:

Hint: Look at how it can fail. We'd want the LHS to be larger than the RHS. We can make the LHS be equal to $|y_1+y_2|$ by just having $x_1=-x_2$. So we can make the LHS as big as we want by, say, choosing $(1,1000)$ and $(-1,1000)$. What happens to the RHS?