I have to prove that for $(x_1,y_1),(x_2,y_2)\in[0,1]^2$ $$|x_1y_1-x_2y_2|\leq|x_1-x_2|+|y_1-y_2|$$ Now what I do is this:

w.l.o.g. say $y_1<y_2$. Def $$f(y)=y(\frac{x_1-x_2}{y_1-y_2}(y-y_2)+x_2)$$ So that $$|x_1y_1-x_2y_2|=|f(y_1)-f(y_2)|$$ So to use the mean value theorem:


So for some $\gamma\in (y_1,y_2)$ \begin{align} |x_1y_1-x_2y_2|&=|(2\gamma\frac{x_1-x_2}{y_1-y_2}-y_2\frac{x_1-x_2}{y_1-y_2}+x_1)(y_1-y_2)|\\ &=|(2\gamma-y_2)(x_1-x_2)+x_1(y_1-y_2)|\\ &\leq|(2\gamma-y_2)||x_1-x_2|+x_1|y_1-y_2|\\ &\leq y_2|x_1-x_2|+|y_1-y_2|\\ &\leq|x_1-x_2|+|y_1-y_2| \end{align}

I think this is correct, but it seems way more tedious then it needs to be. So can someone:

a) verify that this is actually correct

b) Possibly explain how to do this shorter and more directly


1 Answer 1


You can use $\triangle$ -inequality: $|x_1y_1-x_2y_2|= |x_1y_1-x_1y_2 + x_1y_2 -x_2y_2| \leq |x_1y_1-x_1y_2|+|x_1y_2-x_2y_2|=|x_1||y_1-y_2|+|y_2||x_1-x_2|\leq |x_1-x_2|+|y_1-y_2|$

  • $\begingroup$ It seems that your proof is fine, but I still like mine ! $\endgroup$
    – DeepSea
    Apr 3, 2015 at 8:02
  • $\begingroup$ Haha thanks. I completely agree that your proof preferable. $\endgroup$ Apr 3, 2015 at 8:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .