Show that if $a,b \in \mathbb{R}^n$, then $|||a|| - ||b||| \leqslant ||a+b||$ 
Show that if $a,b \in \mathbb{R}^n$, then $$|\|a\| - \|b\|| \leqslant \|a+b\|$$

We have that $$||a|| = ||a+b-b||\leqslant||a+b||+||-b|| = ||a+b||+||b||$$
and that $$||b|| = ||b+a-a||\leqslant||b+a||+||-a|| = ||b+a||+||a||$$
however I don't see how I can continue from here. If i take $||a||-||b||$ I get that $$||a||-||b|| = ||a+b||+||b|| -(||b+a||+||a||) = ||b||-||a||$$
which doesn't help at all. What shold I do here?
 A: $$\|a\| \le \|a + b \| + \|b\|$$
Hence $\|a\|-\|b\| \le \|a+b\|$.
Similarly we have $\|b\|-\|a\| \le \|a+b\|$
Hence $\max(\|a\|-\|b\|, \|b\|-\|a\|) \le \|a+b\|$
That is $|\|a\|-\|b\||\le \|a+b\|$
A: Sneaky trick: write $||a|| = || -a||$, $||a + b|| = ||-a-b||$, and use the triangle inequality directly.
A: As @Siong Thye Goh already did the solution, I'll mention one thing.

$\blacksquare~$ Claim: For any Vector Subspace $(X, \| \cdot \|)$ of $~\mathbb{K}^{n}$, we have the following inequality satisfied.
\begin{align*}
    \| a - b \| \geqslant \big\lvert \| a \| - \| b \| \big
    \rvert  \quad \text{for any } a, b \in X \subseteq \mathbb{K}^{n}
\end{align*}
$\blacksquare~$Proof: We have by $\textbf{triangle inequality of norms}$
\begin{align*}
    &\| (a - b) + b \| ~\leqslant~ \| a - b \| + \| b \|  \quad \text{for any } a, b \in X\\
    \implies &  \| a \| - \| b \|  ~\leqslant~ \| a - b \|  \quad \text{for any } a, b \in X 
\end{align*}
Then $\max(\|a\|-\|b\|, \|b\|-\|a\|) \leqslant \|a-b\|$
Hence, we have that $\left| \| a \| - \|b \| \right| \leqslant \| a - b \|$.

Using the inequality for any $x, x_0 \in X~$ for $(X, \| \cdot \|)$ is a normed linear space and $X$ is a subspace of $\mathbb{R}^n$, we have a very important claim.

$\bullet~$ Claim: The map $\| \cdot \| : X \to [0, \infty)$ is continuous or in other words, the norm $\| \cdot \|$ is continuous.
$\bullet~$ Proof: From the definition of continuity we have, for any given $\epsilon > 0$, there exists $\delta > 0$ such that
\begin{align*}
    \big\lvert \| x \| - \| x_{0} \| \big\rvert < \epsilon ~\text{ when }~ \| x - x_{0} \| < \delta \quad \text{for some arbitrary } x_{0} \in X 
\end{align*}
from the previous problem we have the inequality
\begin{align*}
    \big\lvert \| x \| - \| y \| \big\rvert \leqslant \| x - y \| \quad \text{for any } x, y \in X 
\end{align*}
Let us pick our $\epsilon = \delta$. Therefore we have
\begin{align*}
    \big\lvert \| x \| - \| x_{0} \| \big\rvert \leqslant \| x - x_{0} \| < \delta = \epsilon \quad \text{for some arbitrary } x_{0} \in X
\end{align*}
Which shows that the map $\| \cdot \|$ is continuous at $x_{0}$. As $x_{0}$ is arbitrary, then the function $\| \cdot \|$ is continuous on the whole space $X$.

This makes the important proof of any any norm being continuous on a finite dimensional vector subspace of $\mathbb{K}^n$.
Not related to the question though, no intension to spam :)
