The least value of product xyz for which determinant |x111y111z| is not negative is

|x111y111z|≥0 xyz−x−y−z+2≥0 xyz+2≥x+y+z≥3(xyz)13(∵AM≥GM) xyz+2−3(xyz)13≥0 Let (xyz)=t3 t3−3t+2≥0 (t+2)(t−1)2≥0 ⇒t+2≥0,t≥−2 t3=(−2)3⇒−23

The least value of product xyz for which determinant |x111y111z| is not negative is

|x111y111z|≥0 xyz−x−y−z+2≥0 xyz+2≥x+y+z≥3(xyz)13(∵AM≥GM) xyz+2−3(xyz)13≥0 Let (xyz)=t3 t3−3t+2≥0 (t+2)(t−1)2≥0 ⇒t+2≥0,t≥−2 t3=(−2)3⇒−23

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The least value of product $x y z$ for which determinant $| \begin{matrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{matrix} |$ is not negative is

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$- 2^{3 / 2}$

$- 1$

$- 2^{\frac{9}{2}}$

$- 2^{3}$

answer is D.

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Detailed Solution

$| \begin{matrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{matrix} | \geq 0 x y z - x - y - z + 2 \geq 0 x y z + 2 \geq x + y + z \geq 3 {(x y z)}^{\frac{1}{3}} (∵ A M \geq G M) x y z + 2 - 3 {(x y z)}^{\frac{1}{3}} \geq 0 L e t (x y z) = t^{3} t^{3} - 3 t + 2 \geq 0 (t + 2) {(t - 1)}^{2} \geq 0 \Rightarrow t + 2 \geq 0, t \geq - 2 t^{3} = {(- 2)}^{3} \Rightarrow - 2^{3}$