An algebra problem by أحمد الحلاق

Since the function is always positive, I first change to the equivalent problem of minimizing $|x+1|+|x+2|+|x-3|$ . You could either examine the function on each interval, which will give you linear equations in each case, where you can just check the endpoints of each interval of the piecewise function; or you could take the derivative, which is $\frac{x+1}{|x+1|}+\frac{x+2}{|x+2|}+\frac{x-3}{|x-3|}$ . Looking at the derivative, each fraction is either 1, -1 or undefined. Since three odd numbers cannot add up to the even number 0, simply check the points where the derivative is undefined, i.e., $x=-2,-1,3$ . One finds that the minimum of the three is achieved when $x=-1$ , which leads to the minimum value $5$ . In other words, the sought answer is $\frac{1}{5}$ .