# Prove inequality statements

Given the three inequalities:

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

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.

• 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. – Thomas Andrews Dec 1 '15 at 18:42
• 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. – Thomas Andrews Dec 1 '15 at 18:44

The second inequality is false whenever $c<0$ , $d<0$ and $|ab|\le |cd|$.
The third inequality is clearly false, if $\alpha\ge 0$.
• 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. – user2850514 Dec 1 '15 at 18:47
• Unfortunately, there are values satisfying the given conditions, for which the inequalities $2$ and $3$ are true and values, for which they are false. – Peter Dec 1 '15 at 18:54