Minimum Value of Sum Absolute Functions

Question

I have this problem

So the minimum value of the |5x-1| is the 0 because |-a|=a and then every negative value of the 5x-1 is positive and greater than 0?

Answer 1

Note that $\,|A|+|B|+|C|\ge|A+B+C|=5|x-1|\ge0$. Particularly, $\,5|x-1|=0\,$ for $\,x=1$.

Further $$A(1,y)=1\cdot y+3\cdot1-y-3=0$$ $$B(1,y)=4\cdot 1+y-4-1\cdot y=0$$ $$C(1,y)=2-2\cdot1=0$$ Therefore $$|A(1,y)|+|B(1,y)|+|C(1,y)|=0$$ So, it is proved that your function takes a mininum ($0$) for $\,x=1$.