Inequality bound using endpoints? Currently reading Introduction to Calculus and Analysis by Richard Courant. On page 42 Courant uses the following fact in a larger example explaining why $x^2$ is not uniform continuous (where $a$ and $b$ are the endpoints of the closed interval of interest).
$$|x - x_0| |x + x_0| \space \leq \space 2 \space |x-x_0| \space (|b| + |a|)$$
So my question is why does this work? It is not clear to me why $|x + x_0| \leq 2 \space (|b| + |a|)$ is true.
Not sure if this will help, but I've included the relevant parts of the textbook. Any help would be appreciated.

The function $f(x) = x^2$ for an infinite $x$-interval is definitely not uniform continuous. It is clear that small changes in x can produce arbitrarily large changes in $x^2$ if only $x$ is large enough. A glance at a table of squares of integers $x$ shows how successive squares are spaced further and further apart as $x$ increases. If, however, we only consider pairs of values $x$ and $x_0$ belonging to a fixed finite closed interval $[a, b]$, we can find a uniform modulus of continuity. Indeed, for $|x - x_0| < \delta$ we have
$$|f(x) - f(x_0)| = |x^2 - x_0^2| = |x - x_0| |x + x_0| \space \leq \space 2 \space |x-x_0| \space (|b| + |a|)$$
$$< 2 \delta (|b| + |a|) = \epsilon$$


Now $x,x_0$ are both in the interval $[a,b]$ thus they are both a convex combination of endpoints, so
$$
x=\theta a+(1-\theta)b,\quad x_0=\alpha a+(1-\alpha)b,\quad \theta,\alpha \in [0,1].$$
Now sum $x$ and $x_0$ and apply the triangle inequality to obtain the result. Specifically
$$|x+x_0| \leq |a|(|\alpha|+|\theta|) + |b||2-\theta-\alpha|\leq 2(|a|+|b|)$$
since we know $\theta$ and $\alpha$ are in $[0,1]$ thus
$$(|\alpha|+|\theta|)\leq 2\quad and\quad|2-\alpha-\theta|\leq 2.$$
