## anonymous 4 years ago $\left|\frac{1}{a^2}-\frac{1}{b^2}\right|\leq\frac{2|a-b|}{k^3}$where$|a|,|b|>k>0$

1. anonymous

$\left|\frac{1}{a^2}-\frac{1}{b^2}\right|=\left|\frac{b^2-a^2}{(ab)^2}\right|=\frac{|a^2-b^2|}{(ab)^2}=\frac{|(a-b)(a+b)|}{(ab)^2}\leq\frac{|a-b||a+b|}{k^4}$now, all i have to prove is$|a+b|\leq2k??$

2. anonymous

i dont know where ive gone wrong if at all -_-

3. watchmath

do we know the relation between |a| and |b|?

4. anonymous

it just says $$|a|,|b|>k>0$$ :(

5. anonymous

ah damn i forgot there was a 2 in the numerator...

6. anonymous

:(

7. anonymous

You must make an assumption. Lets do two cases: $C1 \rightarrow |a|>|b|>k; C2 \rightarrow k<|a|<|b||; C3 \rightarrow |a|=|b|>k$ What can we say based off these assumptions? Well for the first one. We would then have that: $\frac{1}{a^2}-\frac{1}{b^2}<0 \implies \left|\frac{1}{a^2}-\frac{1}{b^2} \right|=\frac{1}{b^2}-\frac{1}{a^2}$ By the definition of absolute value. Since they are positive that means that a-b>0 so |a-b|=a-b by def. Leaves us with: $\frac{1}{b^2}-\frac{1}{a^2} \le ^? \frac{2(a-b)}{k^3}$ Since k>0 we can multiply by k^3 and divide by 2 giving: $\left( \frac{1}{b^2}-\frac{1}{a^2} \right)\frac{k^3}{2} \le^?(a-b); \forall |a|>|b|>k>0?$ Getting a common denominator and doing some algebra gives: $\frac{a^2-b^2}{b^2-a^2}\frac{k^3}{2} \le^? (a-b) \implies -\frac{k^3}{2}\le(a-b)$ We KNOW this is true because: $a,b,k>0 \implies (a-b),k^3>0$ So now we have reached a fact that we know is TRUE! Now you must REVERSE the logic and argue why, starting with what we just derived that we know as FACT!!!, and argue why the initial is true. Think about it, you can start with something that is false and derive something true, you must start with something you know is true (the last bit) and use logic to get back up. I hope this helps, the other cases are similar.

8. anonymous

I meant 3 cases ^^^

9. anonymous

The bottom should be a^2b^2 not b^2-a^2, so that changes it a bit...hmmm...lets go from there: $\frac{a^2-b^2}{a^2b^2}\frac{k^3}{2} \le^?(a-b) \implies \frac{(a-b)(a+b)}{a^2b^2}\frac{k^3}{2} \le^? (a-b)$ Divide out the a-b giving (since a>b implies a-b=/=0) $\frac{a+b}{a^2b^2}\frac{k^3}{2} \le^? 1$ $a+b \le^? \frac{2a^2b^2}{k^3}$ I think from here we can argue. Because a,b>0 so and a>b>k>0 so a^2b^2 should always beat out k^3... I'm not sure, its something simliar lol