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Algebra Level 3

If $\lambda$ and $\mu$ be real numbers such that $x^3- \lambda\cdot x^2 + \mu\cdot x - 6 = 0$ has its roots real and positive,then $\text{minimum value}$ of $\mu$ is

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3 \times \sqrt[3]{36}

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Chew-Seong Cheong
Sep 10, 2015

Let the positive real roots of $x^3 - \lambda x^2 + \mu x - 6$ be $a$ , $b$ and $c$ . Using Vieta's formulas, we have:

$\begin{cases} a+b+c & = \lambda \\ ab+bc+ca & = \mu \\ abc & = 6 \end{cases}$

Since $a$ , $b$ and $c$ are positive and real we can use AM-GM inequality:

$\begin{aligned} ab+bc+ca & \ge 3 \sqrt [3] {a^2b^2c^2} \\ \mu & \ge 3\sqrt[3]{6^2} = \boxed{3\times \sqrt[3]{36}} \end{aligned}$

Thanks for the solution!!

Akhil Bansal - 5 years, 9 months ago

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