Prove $|a+b|+|a-b| \geq |a|+|b|$ I am fighting with this proof-writing problem for a while. The statement says $$|a+b|+|a-b| \geq |a|+|b|.$$
I know the triangle inequality which says$$|a+b| \leq |a|+|b|.$$ 
How can I use this inequality to prove the statement above? I am adding the distance between $a$ and $b$ to $|a+b|$ which then turns the triangle inequality to another direction. How can I prove this in a mathematical way? 
 A: Hint: 
If $a$ and $b$ have different signs, then their magnitudes are adding up in the same direction in 
$|a-b|$. Otherwise, i.e. if they have the same signs, this is happening in 
$|a+b|$. And the other term on the LHS ($|a+b|$ in the first case and $|a-b|$ in the second) 
is always positive. It is all in magnitudes when working with absolute values. 
A: Denote
$$ 
{x=a+b \\
y=a-b}
$$
Then, using the triangle inequality,
$$|a|=\dfrac{1}{2}(|x+y|)\leqslant\dfrac{1}{2}(|x|+|y|) \\
|b|=\dfrac{1}{2}(|x-y|)\leqslant\dfrac{1}{2}(|x|+|y|),
$$
hence 
$$
|a|+|b|\leqslant |x|+|y|=|a+b|+|a-b|.
$$
A: I'm following up on what said potato about complex numbers:
Let $a$ and $b$ be two complex numbers. In the complex plane the points with affixes $0$, $a$, $b$ and $a+b$ form a parallelogram. Two diagonals of a parallelogram have a total length higher than the sum of the lentgh of two adjacent sides (use triangular inequality in two adjacent triangles). This means that: $|a-b| + |a+b| \geq |a| + |b|$.
In particular this holds if $a$ and $b$ are real (the parallelogramm is flat then).
This could be generalized to vectorial normed spaces. Hope this helps.
A: $2|a|=|a+b+a-b|\leq |a+b|+|a-b|$
$2|b|=|b+a+b-a|\leq |b+a|+|b-a|=|a+b|+|a-b|$
Add these inequalities and then halve.
A: EDIT: I give a new proof, that is better than the old one (see the old one below).
I am going to prove something stronger: $|a+b|+ |a-b| \ge 2\max \{ |a|,|b| \}$.
Assume WLOG that $|a| \ge |b|$. Square both sides to get the following equivalent inequality (using $|a\pm b|^2 = |a|^2 + |b|^2 \pm 2(a\overline{b} + b\overline{a})$:
$$2(|a|^2+|b|^2) + 2|a^2-b^2| \ge 4|a|^2$$
Which is the same as:
$$|a^2-b^2| \ge |a^2|-|b^2|$$
Which in turn is a direct application of the triangle inequality.
My old proof:
Since both sides are non-negative, if I take the square of both sides I get an equivalent inequality. By using $|a\pm b|^2 = |a|^2 + |b|^2 \pm 2(a\overline{b} + b\overline{a})$, we get:
$$|a|^2 + |b|^2 + 2|a^2-b^2| \ge 2|ab|$$
Since $(|a|-|b|)^2 \ge 0$, we have $|a|^2+|b|^2 \ge 2|ab|$, and because $|a^2-b^2|\ge 0$, the inequality follows.
Equality occurs when $|a^2-b^2|=0$ and $|a|=|b|$, i.e. $a=\pm b$.
(I must admit I prefer ashley's proof, although after my edit my proof works for complex numbers too)
