## anonymous 5 years ago please, please help! how can i show that$\frac{|a+b|}{(ab)^2}\leq\frac{1}{k^3}$when $$a,b,k\in\mathbb{R}$$ and $$|a|,|b|>k>0$$?

1. karatechopper

multiply $ab ^{2}$ on both sides?

2. anonymous

For the denominator: If both a and b are positive, k is definitely greater than the absolute values of both. If one factor is neg. and the other is pos., it still doesn't matter because the product is being squared and will be pos and greater than both in the end. If both a and b are neg., the product is pos. and k is still greater.

3. anonymous

how is k greater than the absolute values of both? It says |a| and |b| are > k.

4. anonymous

oh, oops sorry

5. anonymous

I think this isn’t necessarily true. For example take $$a=b>0$$; Then $$\frac{|a+b|}{(ab)^2}=\frac2{a^3}$$, which is not $$\le\frac1{k^3}$$ for $$a$$ is just a little bigger than $$k$$.

6. anonymous

The problem statement is correct? It's not |a| > |b| > k > 0?

7. anonymous

Nvm. Though I don't think a=b is a counter example though. 2/a^3 should be <= 1/k^3

8. anonymous

Maybe use proof by contradiction. Assume the inequality is false and take a=b > 2*k > 0. Then: (a+a)/(a*a)^2 > 1/k^3 2a/a^4 > 1/k^3 2/a^3 > 1/k^3 Multiply both sides by (1/2). So: 1/a^3 > 1/2k^3 But this isn't true for all |a|,|b|, > k > 0 since 1/a^3 is certainly less than 1/2k^3 in our example which is a contradiction therefore the original equality must be true.

9. anonymous

Sorry a few re-posts to fix typos.

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