Question

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 \(|a|,|b|>k>0\)?

Answer 1

multiply \[ab ^{2}\] on both sides?

Answer 2

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.

Answer 3

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

Answer 4

oh, oops sorry

Answer 5

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 \(k\).

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

Sorry a few re-posts to fix typos.