The least value of the product xyz (such that x, yandz are positive...

Since $A . M \ge G . M$

$\dfrac{x + y + z}{3} \ge \left(xyz\right)\dfrac{1}{3}$

$x + y + z \ge 3\left(xyz\right)\dfrac{1}{3}$

Now, \left|$$\begin{array}{ll}x \\ 1 \\ 1\end{array}$$$$\begin{array}{ll}1 \\ y \\ 1\end{array}$$$$\begin{array}{ll}1 \\ 1 \\ z\end{array}$$\right| = xyz - \left(x + y + z\right) + 2$</p><p>Now determinant is non-negative</p><p>$⇒ xyz - \left(x + y + z\right) + 2 \ge 0$</p><p>$⇒ xyz - \left(3\right)\left(xyz\right)\dfrac{1}{3} + 2 \ge 0$</p><p>Let$xyz = {t}^{3}$</p><p>So,${t}^{3} - 3t + 2 \ge 0$</p><p>$⇒ \left(t + 2\right) ( {t}^{2} - 2t + 1 ) \ge 0 ⇒ \left(t + 2\right){( t - 1 )}^{2} \ge 0$</p><p>$⇒ t \ge - 2$</p><p>$⇒ {t}^{3} \ge - 8$</p><p>$⇒$Least value of$xyz = - 8