Suppose that $a,b$ are reals such that the roots of $ax^3-x^2+bx-1=0$ are all positive real numbers. Prove that:

$(i)~~0\le 3ab\le 1$
$(ii)~~b\ge \sqrt3$.

My attempt:

I could solve the first part by Vieta's theorem. But, I am stuck on the second part. Please help. Thank you.


Let $x, y, z > 0$ be the three roots. Then, $x+y+z = xyz = \dfrac1a$ and $xy+yz+zx = \dfrac{b}a$

$(i),\quad $ Clearly, $a, b > 0$. Also $(x+y+z)^2 \ge 3(xy+yz+zx) \implies \dfrac1{a^2} \ge 3\dfrac{b}a \implies 1 \ge 3ab$.

For $(ii),\quad (xy+yz+zx)^2 \ge 3xyz(x+y+z) \implies \dfrac{b^2}{a^2} \ge 3\dfrac{1}{a^2} \implies b \ge \sqrt3$.

P.S. In case the second inequality used is not familiar, you can show that it is equivalent to the following rearrangement: $$(xy)^2+(yz)^2+(zx)^2 \ge (xy)(yz)+(yz)(zx)+(zx)(xy)$$

  • $\begingroup$ Yes, that's very nice. Thanks. Anyways, I could do the first part. $\endgroup$ – Swadhin Feb 7 '15 at 6:22

Hint: $x_1^3 +x_2^3 + x_3^3 = (x_1+x_2+x_3)^3 - 3(x_1+x_2)(x_2+x_3)(x_3+x_1) \leq (x_1+x_2+x_3)^3 - 24x_1x_2x_3$, by AM-GM inequality, and write:

$b = \dfrac{3+\displaystyle \sum x_i^2-a\displaystyle \sum x_i^3}{\displaystyle \sum x_i}$. Express $b$ as a function of $a$ and find the critical points, and take it from there. Use Vieta's theorem again. Use Cauchy-Schwarz inequality for: $\displaystyle \sum x_i^2 \geq \dfrac{1}{3}\cdot \left(\displaystyle \sum x_i\right)^2$ also. This implies:

$b \geq f(a)$, and you can take it from ...there..


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