# Proving that: $| a + b | + |a-b| \ge|a| + |b|$ [duplicate]

I am trying to prove this for nearly an hour now:

$$\tag{\forall a,b \in \mathbb{R}}| a + b | + |a-b| \ge|a| + |b|$$

I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ? Thanks in advance.

• Over $\mathbb{R}$ or $\mathbb{C}$? Oct 17, 2017 at 16:42
• a and b are elements of the real numbers Oct 17, 2017 at 16:42
• B.t.w., this is an inequality, not an inequation, and it has to be proved, not to be solved. Oct 17, 2017 at 16:50
• Sorry for that :) Oct 17, 2017 at 16:51
• Oct 17, 2017 at 16:54

Using triangle inequality, $$|a+b| + |a-b| \geqslant |(a+b) + (a - b)| = 2|a|$$ also as $|a-b| = |b-a|$, $$|a+b| + |a-b| \geqslant |(a+b) + (b - a)| = 2|b|$$ Now add and conclude!

My suggestions would be the following.

1. First observe that it is symmetric in $a$ and $b$. Moreover, it's certainly true when $a = b$.
2. Note also that if we replace $b$ with $-b$, then the claim is unchanged. So we may assume $b \ge 0$. The same holds for $a$.
3. It also shows us that, without loss of generality, we can assume that $a > b$. If $a > b$, then this tells us something useful: $|a - b| = a - b$.

So we need only consider $a > b \ge 0$. I leave the rest of the argument to you (although there's not much left). Hopefully this helps show how one might approach these questions, not just give you an almost-solution to one particular question :)

• can you explain why we can assume that |a−b|=a−b ? Oct 17, 2017 at 16:47
• Note that the inequality is symmetric in $a$ and $b$: swapping them has no effect. We must have $a \ge b$ or $b \ge a$. Since we can swap $a$ and $b$, without loss of generality we just assume that it's $a$ that is larger. Does that explain it? :) Oct 17, 2017 at 16:49
• Oh yeah..... Now i see it... :) Oct 17, 2017 at 16:53

Start with $|a|+|b|$ and rewrite $a$ and $b$ as $\frac{a+b}{2}+\frac{a-b}{2}$ and $\frac{a+b}{2}+\frac{b-a}{2}$ respectively. Use the triangle inequality.

• In other words, if $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ this is equivalent to $|x+y|+|x-y|\leq 2|x|+2|y|$. Oct 17, 2017 at 16:56

To prove $$| a + b | + |a-b| \ge|a| + |b|$$

Square both the sides. This does not change inequality. We have

$$| a + b |^2 + |a-b|^2 + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|$$

$$(|a|^2 + |b|^2 +2|a||b|cos\theta) + (|a|^2 + |b|^2 -2|a||b|cos\theta) + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|$$ where $\theta$ is angle between a and b

$$2|a|^2 + 2|b|^2 + 2|a+b||a-b| \ge|a|^2 + |b|^2 + 2|a||b|$$

$$|a|^2 + |b|^2 + 2|a+b||a-b| \ge 2|a||b|$$

$$|a|^2 + |b|^2 + 2|a+b||a-b| - 2|a||b| \ge 0$$

$$(|a|-|b|)^2 + 2|a+b||a-b|\ge 0$$

So on left hand side we have both terms which are always greater than 0, hence this inequality always holds

Equality exists when a=b

QED

Here's the way I see this geometrically in $\mathbb{C}$: let's say we have two complex numbers $a, b$ and consider the parallelogram formed by $0, a, a + b, b$. The midpoints of the diagonals coincide at the point $\frac{a + b}{2}$. These diagonals cut the parallelogram into four triangles, on each of which we can perform the triangle inequality. We get the following inequalities:

\begin{align*} |a - 0| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right| \\ |(a + b) - a| &\le \left| (a + b) - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - a \right| \\ |b - (a + b)| &\le \left| b - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - (a + b) \right| \\ |0 - b| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right| \end{align*} Simplifying the above inequalities and summing them up yields the desired inequality.

Without loss of generality, we may assume that $|a|\geq |b|$. Since the terms are all non-negative, by squaring both sides, we obtain the equivalent inequality $$(a + b)^2 + (a-b)^2 +2(a^2-b^2)\ge a^2+b^2+2|a||b|$$ that is $$3a^2-b^2\ge 2|a||b|\Leftrightarrow (3|a|+|b|)(|a|-|b|)\geq 0$$ which holds. Therefore the given inequality is always true.

• How do you know that a >= b? Oct 17, 2017 at 16:48
• We solve the inequality for $a\geq b$. For $b\geq a$, we simply swap $a$ and $b$. Oct 17, 2017 at 16:50

Yet another solution:

\begin{eqnarray} |a+b|+|a-b| &=& \max(a+b,-a-b)+\max(a-b,b-a) \\ &=& \max(2a, 2b, -2b, -2a) \\ &=& 2 \max(|a|,|b|) \\ &\ge& |a|+|b| \end{eqnarray}

From triangular inequality we have

$$\left|u+v\right|+\left|u-v \right|\le |u|+|v|+|u|+|v|=2|u|+2|v|\\ \left|\frac{u+v}{2}\right|+\left|\frac{u-v}{2}\right|\le |u|+|v|\quad(*)$$

set $a+b=u;\;a-b=v$

$$a=\frac{u+v}{2};\;b=\frac{u-v}{2}$$

$| a + b | + |a-b| \ge|a| + |b|$

$$\left|\frac{u+v}{2}+\frac{u-v}{2}\right|+\left|\frac{u+v}{2}-\frac{u-v}{2}\right|\ge \left|\frac{u+v}{2}\right|+\left|\frac{u-v}{2}\right|$$

$$|u|+|v|\ge \left|\frac{u+v}{2}\right|+\left|\frac{u-v}{2}\right|$$ which is true because of $(*)$

Very Simple Trick: We have that \begin{split} (|a|-|b|)^2 +2|a^2-b^2| \ge 0&\Longleftrightarrow & a^2+b^2 -2|a||b|+ 2|a +b||a-b| \ge 0\\ &\Longleftrightarrow & a^2+b^2 + 2|a +b||a-b|\ge 2|a||b|\\ &\Longleftrightarrow& \color{red}{2a^2+2b^2} + 2|a +b||a-b|\ge \color{red}{a^2+b^2}+2|a||b|\\ &\Longleftrightarrow& (|a +b|+|a-b|)^2 \ge (|a|+|b|)^2\\ &\Longleftrightarrow& |a +b|+|a-b| \ge |a|+|b|\end{split}

Given that $$\color{red}{ (|a +b|+|a-b|)^2 = 2a^2+2b^2 + 2|a +b||a-b|}$$