# Inequality involving roots of a third degree polynomial

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 : $2\alpha+\beta \leq a$.

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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

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