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

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• solakis1
In summary, we need to find a value of B that satisfies the conditions for all x, y, a, and b such that the absolute value of the expression |x^2y - a^2b| is less than A. Using the triangle inequality, we can simplify the expression to be less than 3B. Therefore, taking B = 1/3A would satisfy the condition and ensure that |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 $$\displaystyle |x^2y-a^2b|<A$$

solakis said:
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 $$\displaystyle |x^2y-a^2b|<A$$
[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]

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