The least value of the product xyz for which | vedclass

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Question

$\begin{array}{*{20}{c}}
x&1&1\
1&y&1\
1&1&z
\end{array} \ge 0$

$xyz - x - y - z + 2 \ge 0$

$xyz + 2 \ge x + y + z

Vedclass · Accepted Answer

$-8$

Vedclass · Answer

$\begin{array}{*{20}{c}}
x&1&1\
1&y&1\
1&1&z
\end{array} \ge 0$

$xyz - x - y - z + 2 \ge 0$

$xyz + 2 \ge x + y + z \ge 3{\left( {xyz} ight)^{1/3}}$

$xyz + 2 - 3{\left( {xyz} ight)^{1/3}} \ge 0$

at $\left( {xyz} ight) = {t^3}$

${t^3} - 3t + 2 \ge 0$

$\left( {t + 2} ight){\left( {t - 1} ight)^2} \ge 0$

$\left[ {t =  - 2} ight]{t^3} =  - 8$

The least value of the product $xyz$ for which the determinant $\left| {\begin{array}{*{20}{c}}
x&1&1 \\
1&y&1 \\
1&1&z
\end{array}} \right|$ is non-negative, is

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