Does $|d(x_n.a)-d(a,x)|\leq d(x_n,x)$?

Question

Let $(M,d)$ be a metric space and $x\in M$, $x_n\to x$

So $|d(x_n,a)-d(a,x)|\leq d(x_n,x)$ is a concluse form the triangle inequallty?

Answer 1

By triangle inequality, $d(x_n,a)\leq d(x_n,x)+d(x,a)\implies d(x_n,a)-d(a,x)\leq d(x_n,x)~~~~~~~~~~~(1)$.

Next, again by triangle inequality, $d(a,x)\leq d(a,x_n)+d(x_n,x)\implies d(a,x)-d(x_n,a)\leq d(x_n,x)~~~~~~~~~~~(2)$.

From $(1)$ and $(2)$, $$|d(x_n,a)-d(a,x)|\leq d(x_n,x).$$

Answer 2

$$|d(x_n,a)-d(a,x)|\leq d(x_n,x)\iff $$

$$ - d(x_n,x) \le d(x_n,a)-d(a,x)\le d(x_n,x) \iff $$

$$d(a,x)- d(x_n,x) \le d(x_n,a)\le d(x_n,x)+d(a,x)$$

Apply triangle inequality and you are done.