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

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The discussion centers on proving that for a polynomial P(x) with three real roots, at least two of which are distinct, the inequality A²+B²+18C>0 holds. The roots are represented as a, b, and c, leading to the expressions for A, B, and C based on these roots. The use of the AM-GM inequality demonstrates that the sum of the roots and the sum of the products of the roots are both greater than zero under certain conditions. However, a counterexample is presented where two roots are negative, resulting in A²+B²+18C being less than zero, suggesting that an additional condition is necessary to ensure the inequality holds. The conclusion emphasizes that the problem's validity hinges on excluding cases where two roots are negative.
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If a polynomial $P(x)=x^3+Ax^2+Bx+C$ has three real roots at least two of which are distinct, prove that $A^2+B^2+18C>0$.
 
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anemone said:
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 $$(a+b+c)^2 + (bc+ca+ab)^2 + 18abc > 9m^2 + 9m^4 + 18m^3 = 9m^2(1-m)^2 \geqslant0.$$
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Opalg said:
[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

Cannot apply AM-GM inequality as a,b,c are not positive
 
kaliprasad said:
Hello Opalg

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.

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Opalg said:
[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.
 
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