Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b$ be two positive numbers such that $a^3 \gt 27b$. Consider the polynomial

$$ W(x)=x^3-2ax^2+a^2x-4b $$

Then we have

$$ W(0)=-4b \lt 0, \ W(\frac{a}{3})=\frac{4}{27}(a^3-27b) \gt 0, \ W(a)=-4b \lt 0 $$

We deduce that $W$ has three roots $\alpha,\beta,\gamma$ with

$$ 0 \lt \alpha \lt \frac{a}{3} \lt \beta \lt a \lt \gamma $$

Prove or find a counterexample : $2\alpha+\beta \leq a$.

share|cite|improve this question
What happens when you work out $W(2\alpha+\beta)$? – Gerry Myerson Jan 1 '13 at 13:48
@GerryMyerson : $W(2\alpha+\beta)$ is exactly $-6a^2\alpha + (8\alpha^2 + 4\beta\alpha)a + (8b + 6\beta\alpha^2)$, an expression whose sign is not obvious. So what ? – Ewan Delanoy Jan 1 '13 at 14:05
Sorry, just thought it might be worth a try. – Gerry Myerson Jan 1 '13 at 23:29
up vote 4 down vote accepted

$W(\alpha + a)=a\, \alpha\,(3 \alpha-a)\leq 0$ so $\alpha + a \leq \gamma$. Together with $\alpha+\beta+\gamma=2a$ this implies that $2\alpha + \beta = 2a +\alpha-\gamma\leq a$.

share|cite|improve this answer
Right. May I ask if you found this simple answer by some sort of systematic method ? Because I tried proofs along similar lines and failed, always encountering expressions whose sign was not obvious. – Ewan Delanoy Jan 1 '13 at 14:04
@EwanDelanoy Quasi systematic at best. In this case I started with the last inequality and concluded that $\gamma-\alpha \geq a$ would suffice. Then I tried $W(\alpha + a)$ and got lucky. – WimC Jan 1 '13 at 15:30
by the way, I also the need the "reverse" inequality $a \leq \alpha+2\beta$. Because of your humiliatingly simple solution, I’ll try harder to find a proof for myself before asking it officially here. But if once again, a one-line proof leaps to your lucky eye, let me know, it might save me some trouble ... – Ewan Delanoy Jan 1 '13 at 15:48
@EwanDelanoy Note that $W(4a/3)>0$ so $\beta-\gamma\geq -a$. – WimC Jan 1 '13 at 16:00
How do you deduce $\beta - \gamma \geq -a$ from $W(\frac{4a}{3}) \gt 0$ ? – Ewan Delanoy Jan 1 '13 at 16:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.