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

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  1. karatechopper
    • 5 years ago
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    multiply \[ab ^{2}\] on both sides?

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    how is k greater than the absolute values of both? It says |a| and |b| are > k.

  4. anonymous
    • 5 years ago
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    oh, oops sorry

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    The problem statement is correct? It's not |a| > |b| > k > 0?

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

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Sorry a few re-posts to fix typos.

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