How to solve this problem on absolute value function?

Question

If $a,b\in \mathbb R$ and be distinct numbers satisfying $$|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$$ then the minimum value of $|a-b|$ is ? ($|...|$ represents absolute value)

I tried solving the equalities but couldn't get the value of $a,b$ for which this equality comes true. Please help!

Answer 1

Well, how are we even going to have $a-b$ in this equation? Well, if we look at $\lvert a \rvert+\lvert b \rvert$, we want $\lvert a \rvert=a$ and $\lvert b \rvert=-b$ so that the sum will be $a-b$. Thus, we'll get $a-b$ if the sign of $a$ is positive and the sign of $b$ is negative, giving us: $$\lvert a \rvert+\lvert b \rvert=a+(-b)=a-b$$

Similarly, if $a+1 \geq 0$ and $b+1 \leq 0$, we have $\lvert a+1 \rvert+\lvert b+1 \rvert=(a+1)-(b+1)=a-b$ and if $a-1 \geq 0$ and $b+1 \leq 0$, we have $\lvert a-1 \rvert+\lvert b-1 \rvert=(a-1)-(b-1)=a-b$.

Therefore, we can get conditions where all of these expressions are just $a-b$:

• $a+1 \geq 0$ and $a \geq 0$ and $a-1 \geq 0$ just means $a \geq 1$.
• $b+1 \leq 0$ and $b \leq 0$ and $b-1 \leq 0$ just means $b \leq -1$.

Thus, if $a \geq 1$ and $b \leq -1$, we have a solution because all of the expressions in the equation are just $a-b$! In order to minimize $\lvert a-b \rvert$, we have to get $a$ and $b$ as close to each other as possible. This means $a$ needs to be its least value, which is $a=1$, and $b$ needs to be its greatest value, which is $b=-1$. Thus, we have $\lvert a-b \rvert=\lvert 1-(-1) \rvert=2$.

Now, this is my tentative solution. I haven't proven yet that there is another solution where $a \in [-1, 1]$ or $b \in [-1, 1]$. Therefore, the minimum $\lvert a-b \rvert$ is at most $2$, but I haven't proven tht $2$ is the minimum.

Answer 2

In the plane $ab$, the locus of $|a-i|+|b-j|=k$ is a diamond shape centered at $(i,j)$, with a size growing with $k$.

Here we have three such diamonds, centered at $(-1,-1),(0,0)$ and $(1,1)$. The sides orthogonal to the main bissectrix can never have common points. The other sides start overlapping when the extreme squares meet, i.e. when $k=2$. The two common points are $(1,-1)$ and $(-1,1)$, corresponding to the absolute difference $2$.

Answer 3

The problem can be posed as minimizing $\phi(x)=|x_1-x_2|$ on the set $A=\{ x | \|x-(1,1)\|_1 = \|x\|_1 = \|x-(-1,-1)\|_1 \}$.

Let $\eta(x) = |x_1+x_2|$ and note that $\|\eta(x)| \le \|x\|_1$, $\|\phi(x)| \le \|x\|_1$ and $\|x\|_1 = \max(\eta(x), \phi(x))$. Also note that $\phi(x-(t,t)) = \phi(x)$ for all $t$.

Note that if $\|x-(1,1)\|_1 < 2$ then $|\eta(x)-2 | <2$ and so $\eta(x) >0$, and similarly if $\|x-(-1,-1)\|_1 < 2$ then $|\eta(x)+2 | <2$ and so $\eta(x) <0$.

Hence if $x \in A$ then $\|x\|_1 \ge 2 $. Note that $(1,-1) \in A$, and so $\min_{x \in A} \phi(x) \le \phi((1,-1)) = 2$.

Now suppose $x \in A$ and $\phi(x) < 2$, since $\|x\|_1 \ge 2$ we have $\|x\|_1 = \eta(x)$. Since $\phi(x-(1,1)) = \phi(x-(-1,-1)) = \phi(x)$, and $x \in A$, we have $\|x-(1,1)\|_1 = \eta(x-(1,1)) = \|x-(-1,-1)\|_1 = \eta(x-(-1,-1)) = \|x\|_1$.

Hence $|x_1+x_2 -2| = |x_1+x_2 +2 | = |x_1+x_2|$ which is impossible (since $|a+b| = |a|$ iff $b \in \{0,-2a\}$).

Hence $\phi(x) \ge 2$ for all $x \in A$ and so $\min_{x \in A} \phi(x) = 2$.