If begin{vmatrix} a & 1 & 1 1 & b & 1 1 & 1 | vedclass

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(B) Given the determinant: $\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} > 0$ Expanding along the first row: $a(bc - 1) - 1(c - 1

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$-8$

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(B) Given the determinant: $\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} > 0$ Expanding along the first row: $a(bc - 1) - 1(c - 1) + 1(1 - b) > 0$ $abc - a - c + 1 + 1 - b > 0$ $abc + 2 > a + b + c$ . . . . . . $(i)$ By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$: $\frac{a + b + c}{3} \geq (abc)^{1/3} \Rightarrow a + b + c \geq 3(abc)^{1/3}$ . . . . . . $(ii)$ Substituting $(ii)$ into $(i)$: $abc + 2 > 3(abc)^{1/3}$ Let $x = (abc)^{1/3}$,then $x^3 + 2 > 3x$ $x^3 - 3x + 2 > 0$ $(x - 1)^2(x + 2) > 0$ Since $(x - 1)^2$ is always non-negative,for the inequality to hold,we must have $x + 2 > 0$ $x > -2 \Rightarrow (abc)^{1/3} > -2$ Cubing both sides: $abc > -8$

If $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} > 0$,then $abc >$

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