High school inequality 2 |x^2y-a^2b|<A

[solakis1]

Given A>0, find a B>0 such that :

for all x,y,a,b : if 0<x<1,0<y<1,0<a<1,0<b<1,and |x-a|<B,|y-b|<B,then $$|x^2y-a^2b|<A$$

 

[Opalg]

Given A>0, find a B>0 such that :

for all x,y,a,b : if 0<x<1, 0<y<1, 0<a<1, 0<b<1, and |x-a|<B, |y-b|<B, then $$|x^2y-a^2b|<A$$

[sp]First step: $|x^2-a^2| = |(x+a)(x-a)| = |x+a|x-a| \leqslant 2|x-a|$.

Next (using the triangle inequality), $|x^2y - a^2b| = |x^2y-a^2y + a^2y - a^2b| \leqslant |x^2y-a^2y| + |a^2y - a^2b| = |x^2-a^2|y + a^2|y-b| \leqslant 2|x-a| + |y-b| < 3B.$

So take $B = \frac13A$. Then $|x^2y - a^2b| < 3B = A$.[/sp]