For which values of a does |x - 2| + |x + 3| + |x - 1| + |x + 1| >= a hold for every real x?

Step-by-step solution

1. 1

   Reduce the question to a minimum. The inequality holds for all $x$ exactly when $a$ is no larger than the smallest value the left side ever takes. So define

   $f(x)=|x-2|+|x+3|+|x-1|+|x+1|$

   and find $\min f$. Everything else is bookkeeping.

2. 2

   Use the median shortcut to predict the answer. A sum of absolute deviations $\sum|x-c_{i}|$ is minimised at any median of the points $c_{i}$. Here the four points are $-3,\,-1,\,1,\,2$; with an even count, every $x$ in the middle interval $[-1,1]$ is a median. So the minimum is attained on all of $[-1,1]$ — not on $[-3,1]$, a claim that is easy to make and wrong.

3. 3

   Confirm by splitting into intervals. The breakpoints $-3,-1,1,2$ cut the line into five pieces, and on each one every absolute value opens with a fixed sign:

   $x\le-3:\ f=-4x-1;\qquad -3\le x\le-1:\ f=5-2x;$

   $-1\le x\le 1:\ f=7;\qquad 1\le x\le 2:\ f=2x+5;\qquad x\ge 2:\ f=4x+1.$

   On $[-3,-1]$ the function is the decreasing line $5-2x$, running from $f(-3)=5+6=11$ down to $f(-1)=5+2=7$ — it is not constant there.

4. 4

   Read off the minimum. The two outer pieces increase away from the centre, the two middle-adjacent pieces slope down toward $[-1,1]$, and on $[-1,1]$ the function is the constant

   $f(x)=(2-x)+(x+3)+(1-x)+(x+1)=7.$

   So $\min f=7$, attained exactly on $[-1,1]$. Spot checks: $f(0)=2+3+1+1=7$ ✓, $f(-2)=4+1+3+1=9$ ✓ (strictly above the minimum), $f(-3)=5+0+4+2=11$ ✓.

5. 5

   State the condition on $a$. The inequality $f(x)\ge a$ holds for every real $x$ if and only if

   $a\le 7.$

   If $a>7$ the inequality fails on the whole open middle band, and its solution set is then $\left(-\infty,\tfrac{5-a}{2}\right]\cup\left[\tfrac{a-5}{2},\infty\right)$ for $7<a\le 9$, obtained by inverting the two adjacent linear pieces.