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?


$2|a|=|a+b+a-b|\leq |a+b|+|a-b|$

$2|b|=|b+a+b-a|\leq |b+a|+|b-a|=|a+b|+|a-b|$

Add these inequalities and then halve.

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    $\begingroup$ great, short and sharp! thanks :D $\endgroup$ – doniyor Dec 14 '12 at 23:45

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.

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    $\begingroup$ great, thanks, nice guidance for future! $\endgroup$ – doniyor Dec 14 '12 at 23:31
  • $\begingroup$ Yw. glad that helps. $\endgroup$ – ashley Dec 14 '12 at 23:33
  • $\begingroup$ This doesn't work if $a$ and $b$ are assumed to be complex numbers (and the statement certainly holds in that case). Actually it holds in any metric space... $\endgroup$ – Potato Dec 14 '12 at 23:40
  • $\begingroup$ @Potato: that's out of context and irrelevant here. $\endgroup$ – ashley Dec 15 '12 at 0:22
  • $\begingroup$ Not really. I admit the metric space comment is a bit of a stretch, but this could easily be a problem about complex numbers. And it's good to know which methods generalize and which don't. (But this is a good and commonsense way to go about it.) $\endgroup$ – Potato Dec 15 '12 at 3:02

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 $$ |a|+|b|\leqslant |x|+|y|=|a+b|+|a-b|. $$


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.

  • $\begingroup$ You geometric proof acutally shows $|a-b|+|a+b|\ge 2 \max{|a|,|b|}$ - the diagonals form 4 edges, and you can interpret them in two ways - once as 2 pairs of edges of a triangle, each in front of a side of length $|a|$, or 2 pairs of edges of a triangle, each in front of a side of length $|b|$. $\endgroup$ – Ofir Dec 15 '12 at 11:57

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)


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