Show that $|a| + |b| + |c| \leq |a - |b - c|| + |b - |c - a|| + |c - |a - b||$ where $a, b, c \in \mathbb{R}$ and $a + b + c = 0$

Question

How can we prove that $|a| + |b| + |c| \leq |a - |b - c|| + |b - |c - a|| + |c - |a - b||$ where $a, b, c \in \mathbb{R}$ and $a + b + c = 0$

At first, I thought I should use applied triangle inequality

$$|x_1 + x_2 + ... + x_n| \leq \sum_{i = 1}^{n} |x_i|$$

But in fact, the sign is different, so I think I should prove by other theorems. Can somebody please help me with this?

Answer 1

Since $a+b+c=0$ then we can assume, by relabeling the variables, either $c\le 0\le a\le b$ or $b\le a\le0\le c$.

Assume $c\le0\le a\le b$.

Then the inequality to be proved reduces to $$a+b+|c|\le(b+|c|-a)+(b+|c|+a)+(b+|c|-a)$$ $$2a+2b\le 2a+6b$$

Can you work out the other possibility in the same way?

Answer 2

We need to prove that: $$|a-|a+2b||+|b-|2a+b||+|a+b+|a-b||\geq|a|+|b|+|a+b|.$$ Now, if $b=0$ it's obvious.

Let $b\neq0$ and $a=xb$.

Thus, we need to prove that $$|x-|x+2||+||2x+1|-1|+|x+1+|x-1||\geq|x|+1+|x+1|,$$ which is smooth.

For example, for $x\geq1$ we need to prove that: $$4x+2\geq2x+2,$$ which is obvious.

Answer 3

Another way:

Since our inequality is symmetric, we can assume that $a\geq b\geq c$.

Thus, by the triangle inequality $$\sum_{cyc}|a-|b-c||=|a+c-b|+2|b+c-a|=2|b|+4|a|\geq$$ $$\geq2|b|+2|a|\geq|a|+|b|+|a+b|=\sum_{cyc}|a|.$$

Answer 4

Let’s use $a+b+c=0$

$$ \begin{align} \left|a-\left|b-c\right|\right|&=\left|-b-c-\left|b-c\right|\right|\\ &=\left|b+c+\left|b-c\right|\right|\\ &=2\left|\max{\left(b,c\right)}\right|\\ \\ \left|b-\left|c-a\right|\right|&=2\left|\max{\left(c,a\right)}\right|\\ \left|c-\left|a-b\right|\right|&=2\left|\max{\left(a,b\right)}\right| \end{align} $$

Now combine this result with triangle inequality to obtain

$$ \sum_{cyc}{\left|a-\left|b-c\right|\right|}\geq|a|+|b|+|c|+2\left|\max{\left(a,b,c\right)}\right| $$

Equality occurs when the two biggest numbers are non - negative.

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Extra: with similar steps we can obtain

$$ \sum_{cyc}{\left|a+\left|b-c\right|\right|}\geq|a|+|b|+|c|+2\left|\min{\left(a,b,c\right)}\right| $$

Equality occurs when the two smallest numbers are non - positive.