Absolute value distance Let $a$ and $b$ be in the interval $[0, 10]$, and let $a_0$ have a distance inferior to $10^{-3}$ from $a$ and $b_0$ a distance inferior to $10^{-3}$ from $b$. How close is $a_0 b_0$ from $ab$?
This has got me stuck. How can I verify this through inequalities?

Partial answer: Upper bound for $a_0b_0-ab.$  
For $a\leq 10-10^{-3}$ and $b\leq 10-10^{-3}$ we have $$a_0b_0<(a+10^{-3})(b+10^{-3})=ab+10^{-3}(a+b)+10^{-6}\leq$$ $$\leq ab+10^{-3}(20-2\cdot 10^{-3})+10^{-6}=$$ $$=ab+10^{-3}(20-10^{-3}).$$  Observe that in this case $a_0b_0-ab$ can be arbitrarily close to $10^{-3}(20-10^{-3}).$
For $a\leq 10-10^{-3}$ and $10-10^{-3}<b\leq 10$ we have $$a_0b_0<(a+10^{-3})10=(a+10^{-3})(b+(10-b))=$$ $$= ab+a(10-b)+10^{-3}(10)< ab+a(10^{-3})+10^{-3}(10)\leq $$ $$\leq ab+(10-10^{-3})10^{-3}+10^{-3}(10)=$$ $$=ab+10^{-3}(20-10^{-3}).$$For $10-10^{-3}<a<10$ and $10-10^{-3}<b\leq 10$ we have $$a_0b_0\leq 10^2=(a+(10-a))(b+10-b)=$$ $$=ab +a(10-b)+b(10-a)+(10-a)(10-b)<$$ $$<ab+10(10^{-3})+10(10^{-3})+10^{-6}=ab+10^{-3}(20-10^{-3}).$$
