Roots of a Polynomial Function A²+B²+18C>0

[anemone]

If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.

 

[Opalg]

If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.

[sp]Let $a,b,c$ be the three (real) roots of $P(x)$. Then $A = -(a+b+c)$, $B = bc+ca+ab$ and $C = -abc$. So we want to prove that $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 0.$$

Let $m = (abc)^{1/3}$ be the geometric mean of $a,b,c$. Since those numbers are not all equal, the AM-GM inequality is strict, so that $a+b+c > 3m$. For the same reason, $bc+ca+ab > 3m^2$. Therefore $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 9m^2 + 9m^4 + 18m^3 = 9m^2(1-m)^2 \geqslant0.$$
[/sp]

 

[kaliprasad]

[sp]Let $a,b,c$ be the three (real) roots of $P(x)$. Then $A = -(a+b+c)$, $B = bc+ca+ab$ and $C = -abc$. So we want to prove that $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 0.$$

Let $m = (abc)^{1/3}$ be the geometric mean of $a,b,c$. Since those numbers are not all equal, the AM-GM inequality is strict, so that $a+b+c > 3m$. For the same reason, $bc+ca+ab > 3m^2$. Therefore $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 9m^2 + 9m^4 + 18m^3 = 9m^2(1-m)^2 \geqslant0.$$
[/sp]

Hello Opalg

Spoiler

Cannot apply AM-GM inequality as a,b,c are not positive

 

[Opalg]

Hello Opalg

Spoiler

Cannot apply AM-GM inequality as a,b,c are not positive

[sp]Good point – I completely overlooked that. However, if one or three of the roots are negative then $C$ will be positive, so the inequality $A^2+B^2+18C>0$ will certainly hold. So the remaining case to deal with is if two of the roots are negative and the third one is positive. I'll have to think about that ... .

[/sp]
Edit:
[sp]The polynomial $x^3 + x^2 - x - 1 = (x+1)^2(x-1)$ has $A=1$, $B=C=-1$, and $A^2 + B^2 + 18C = -16 <0$. So I think that the problem probably needed an extra condition to exclude the case where two of the roots are negative.

[/sp]

 

[anemone]

[sp]
Edit:
[sp]The polynomial $x^3 + x^2 - x - 1 = (x+1)^2(x-1)$ has $A=1$, $B=C=-1$, and $A^2 + B^2 + 18C = -16 <0$. So I think that the problem probably needed an extra condition to exclude the case where two of the roots are negative.

[/sp]

I just checked the source of the problem, I didn't leave out anything. But you made the point, Opalg, that one such counterexample is suffice to disprove the validity of the problem. The problem is only valid if the condition to exclude the case where two of the real roots are negative is in place.