If the roots of the equation $x^{3} - a x^{2} + 4 x - 8 = 0$ are real and positive, then the minimum value of $a$ is

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Solution

Let $α, β, γ$ be the roots of the given equation,

Then, $α + β + γ = a, α β + α γ + β γ = 4, α β γ = 8$

Since $A . M . \geq G . M . \Rightarrow \frac{α + β + γ}{3} \geq \sqrt[3]{α β γ}$

$\Rightarrow \frac{a}{3} \geq \sqrt[3]{8} \Rightarrow a \geq 6$

$∴$ The minimum value of $a = 6.$

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