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Can someone give me an idea how to show that the real part of the solutions of the equation $$ x^3+bx^2+xc+d$$ arenegative iff $b>0,d>0$ and $-bc<-d$ ?

This question is related to proving that a $3\times 3$ matrix, whose characteristic polynomial is the above, is stable.

I tried using Vieta's formulas but they got me nowhere...

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This is called "the Routh-Hurwitz criterion". For a proof, see p.78 (Theorem 11) of this book. – pharmine Nov 20 '11 at 20:50
up vote 1 down vote accepted

The real parts of the roots are negative iff the image of the imaginary axis under $p(x)=x^3+bx^2+cx+d$ makes $3/2$ turns around $0$ (to see it, complete a big piece of the imaginary axis to a closed curve using a half-circle on the left, and use the argument principle). This means that the image must cross the axes at least $5$ times; as $p(it)=i(-t^3+ct)-bt^2+d$, so $-t^3+ct$ must have $3$ real roots and $-bt^2+d$ two real roots. For the moment it gives $c>0$, $bd>0$. Since we can't have more than these $3+2=5$ intersections with the axes, the intersections must be ordered as $oxoxo$, where $o$'s are the roots of $-t^3+ct$ and $x$'s the roots of $-bt^2+d$ (otherwise we wouldn't go around $3/2$-times). This gives the inequality $\sqrt{c}>\sqrt{d/b}$, i.e. $c>d/b$. We also need $-bt^2+d$ negative for large $t$'s (for the same reason); this gives $b>0$. We thus got the conditions $b,c,d>0$, $c>d/b$, which is equivalent to your conditions.

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