1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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: $$(xy)^2+(yz)^2+(zx)^2 \ge (xy)(yz)+(yz)(zx)+(zx)(xy)$$

$\endgroup$
  • $\begingroup$ Yes, that's very nice. Thanks. Anyways, I could do the first part. $\endgroup$ – Swadhin Feb 7 '15 at 6:22
0
$\begingroup$

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..

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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