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Prove the cubic $$ x^3 - \frac{a+a^2+a^4+b+b^2+b^4}2 x^2 + \frac{a^3+a^5+a^6+b^3+b^5+b^6}2 x - \frac{a^7+b^7}2 $$ has all real roots given that $0 < \sqrt a \leq b \leq a^2$.

I don't see how you would do this cleanly, my only thoughts were using cubic discriminant and checking if it is always nonnegative but this turned out to be too ugly and I was unsure of how to implement the condition.

Another thought i had was if I let the cubic be $x^3 - Bx^2 + Cx - D$, we can individually prove $18BCD \geq 27D^2$ and $B^2C^2 \geq 4(C^3+B^3D)$, and add up the inequalities to prove the cubic discriminant was nonzero.

I played around with smaller values and found that for $b=a^2$, the solutions to the cubic are $a^2, a^4, \frac{a^8+a}{2}$.

Could someone give some starting hints or a solution on a cleaner method?

Jyrki Lahtonen
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  • Please don't deface your old questions like that. It is against the site rules. Just ask another one. – Jyrki Lahtonen Oct 25 '23 at 03:38
  • New askers don't alsways appreciate the fact that the posts on this site are meant to be permanent. In other words, they are also meant to be read by other visitors in years to come. That is why we are strict about not erasing the questions. If you read the fine print, it said that by posting you grant the site the right to decide on the future of the post. – Jyrki Lahtonen Oct 25 '23 at 03:43

1 Answers1

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The OP's observation about the zeros at the extremal point $b=a^2$ (and, by $a\leftrightarrow b$ symmetry hence also at the other extreme $a=b^2$) are a big hint. By expanding and staring at what remains, we find $$ f(a^2)=\frac12(a^2-b)(a^2-b^2)(a-b^2)(a+b^2) $$ and, by symmetry again, $$ f(b^2)=\frac12(b^2-a)(b^2-a^2)(b-a^2)(b+a^2). $$ Given the known inequalities $a^2>b$ and $b^2>a$ we see that $f(a^2)$ and $f(b^2)$ have opposite signs (which way they go depends on the sign of $a^2-b^2$). Therefore there exists a zero between $a^2$ and $b^2$.

Similarly $$f(a^4)=\frac12(a^4-b)(a^4-b^2)(a^4-b^4)$$ and $$f(b^4)=\frac12(b^4-a)(b^4-a^2)(b^4-a^4)$$ have opposite signs, and there is another zero between $a^4$ and $b^4$. A cubic with real coefficients and at least two real zeros obviously has only real zeros.

Leaving it to you to verify that the two ranges don't overlap so that it is impossible for the two zeros above to coincide unless we are in a very special circumstance (with $a=b=1$ the polynomial has a triple zero etc).

Jyrki Lahtonen
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