Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient and have all real roots . Prove that $ab \leq 0$

Question

Let $P (x )=x^4+ax^3+bx+c=0$ and have real coefficient and have all real roots . Prove that $ab \leq 0$

First Let the roots of this polynomial (call it P(x)) be $q,r,s,t$ By Vieta's, $a=-(q+r+s+t)$ and $b=-(qrs+rst+qrt+qst)$ So $ab=q^2(rs+st+rt)+r^2(qs+qt+st)+s^2(qr+qt+rt)+t^2(qr+qs+rs)+4qrst$ But how to find contradiction or use inequality to help ?

Thanks

Answer 1

Let $p$, $q$, $r$ and $s$ be roots of our equation and $$p+q+r+s=4u,$$ $$pq+pr+ps+qr+qs+rs=6v^2,$$ $$pqr+pqs+prs+qrs=4w^3$$ and $$pqrs=t^4,$$ where $v^2$ and $t^4$ can be negative.

Thus, $p$, $q$, $r$ and $s$ are roots of the equation: $$x^4-4ux^4+6v^2x^2-4w^3x+t^4=0,$$ which says that by the Rolle's theorem the equation $$(x^4-4ux^4+6v^2x^2-4w^3x+t^4)'=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ has three real roots.

Now, if $w^3=0$, so we are done because in this case $b=0$.

But for $w^3\neq0$ we see that the equation $$w^3x^3-3v^2x^2+3ux-1=0$$ has three real roots, which says that the equation $$(w^3x^3-3v^2x^2+3ux-1)'=0$$ or $$w^3x^2-2v^2x+u=0$$ has two real roots, which says $$v^4-uw^3\geq0.$$ But in our case $v^2=0$.

Id est, $$uw^3\leq0$$ or $$ab\leq0$$ and we are done!

Answer 2

Suppose $ab>0$. Then

Case I: $a>0$ and $b>0$:

If $c>0$, then $P(x)$ has no sign change and so by Descarte's rule of sign change, $P(x)$ has no real positive roots. However $P(-x)$ has two sign changes and so $P(x)$ must have two or less number of real negative roots. But this contradicts that all four roots are real. Similar contradiction is obtained when $c<0$.

Case II: $a<0$ and $b<0$:

If $c>0$, then $P(x)$ has two sign changes and so by Descarte's rule of sign change, $P(x)$ has two or less real positive roots. However $P(-x)$ has no sign changes and so $P(x)$ has no real negative roots. But this contradicts that all four roots are real. Similar contradiction is obtained when $c<0$.