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I would like to show that $$a+b<0\bigwedge ab>c \implies a+b\pm\mathrm{Re}\left[\sqrt{(a+b)^2-4(ab-c)}\right]<0$$ for $a,b,c\in\mathbb{R}.$

Here is my proof:

\begin{align*} a+b<0 \bigwedge ab>c &\implies a+b\pm \mathrm{Re}\left[\sqrt{(a+b)^2-4(ab-c)}<0 \,\right] \\[-5pt] \Updownarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad&\\[-5pt] a+b<0 \bigwedge ab>c &\implies|a+b|>\sqrt{(a+b)^2-4(ab-c)}, \end{align*}

which is true since $4(ab-c)>0.$ Here it has been used that if the argument of the square root is negative, it becomes purely imaginary and hence $a+b\pm0<0$ is true.

Is this a good way to write this? My concern is that I've used the inequalities on the left to manipulate and simplify the one on the right. Is this okay?

Thanks.

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    what kind of numbers are $$a,b,c$$? – 2017-01-13
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    @Dr.SonnhardGraubner Reals, as stated. Or did you mean w.r.t. some other property? – 2017-01-13

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if we squaring the last inequality we get $$0>-4(ab-c)$$ which is true since $$ab-c>0$$ which is true.

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    Thank you. Yes, I think that is the argument I'm making (otherwise I couldn't invoke the triangle inequality). However, my question is more concerning the notation, i.e., whether or not I can manipulate the inequality on the right using the ones on the left and use this as a proof. I'm sorry if the question is not clear in this regard. – 2017-01-13
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    Btw, I have edited the question to use your version instead of trying to use the triangle inequality, so thank you for that. – 2017-01-13