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Given the three inequalities:

\begin{align} a&<0\\ b&<0\\ c+d&<0 \end{align}

Are the conditions below satisfied? Justify your answer.

  1. $a+b<0$
  2. $ab-cd>0$
  3. $\alpha a + b>0$
  4. $(\alpha a + b)^2 > 4\alpha(ab-cd)$

where $\alpha\not=1$.


  1. True by adding the first two given inequalities.
  2. From the first two inequalities we have that $ab>0$ now I'm not really sure how to include $c$ and $d$ into this equation to form a justification or counterexample.

Note I do not want numerical counterexamples for this! I need a general proof that at least one of these is never correct.

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  • $\begingroup$ Numerical counter-examples is how you disprove stuff. How else would you disprove these? I suppose you could come up with general counter-examples, but they are hardly different from specific numerical counter-eaxmples. $\endgroup$ – Thomas Andrews Dec 1 '15 at 18:42
  • $\begingroup$ For example, we could say $b=a,c=d=2a$ is a counter-example (given $a<0$) for $2$. But that is hardly different from $a=b=-1$ and $c=d=-2$ as a numerical counterexample. $\endgroup$ – Thomas Andrews Dec 1 '15 at 18:44
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The second inequality is false whenever $c<0$ , $d<0$ and $|ab|\le |cd|$.

The third inequality is clearly false, if $\alpha\ge 0$.

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  • $\begingroup$ How have you found that $c<0$ or $d<0$? All we know is that $c+d<0$. I need a general proof not a specific counter example. $\endgroup$ – user2850514 Dec 1 '15 at 18:47
  • $\begingroup$ My interpretation was : Are the inequalities satisfied if the given conditions are satisfied ? And what would be a disproof you would accept ? $\endgroup$ – Peter Dec 1 '15 at 18:49
  • $\begingroup$ A disproof shows that one of the inequalities (1-4) are never satisfied at all given the above 3 inequalities. $\endgroup$ – user2850514 Dec 1 '15 at 18:52
  • $\begingroup$ Unfortunately, there are values satisfying the given conditions, for which the inequalities $2$ and $3$ are true and values, for which they are false. $\endgroup$ – Peter Dec 1 '15 at 18:54
  • $\begingroup$ hmm.. Ok, thank you for the help. $\endgroup$ – user2850514 Dec 1 '15 at 18:57

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