The least value of the product xyz for which the determinant |x 1 1| |1 y 1| |1 1 z| is non-negative, is :

Let the given determinant be
Δ = |x 1 1|
|1 y 1|
|1 1 z|
Expanding along the first row, we have
Δ = x(yz - 1) - 1(z - 1) + 1(1 - y)
= xyz - x - z + 1 + 1 - y
= xyz - x - y - z + 2
Given that Δ ≥ 0
xyz - x - y - z + 2 ≥ 0
xyz ≥ x + y + z - 2
Let x = 2, y = 2, z = 2
Then xyz = 8
x + y + z - 2 = 4
8 ≥ 4
Let x = 1, y = 1, z = 1
Then xyz = 1
x + y + z - 2 = 1
1 ≥ 1
Let x = -1, y = -1, z = -1
Then xyz = -1
x + y + z - 2 = -5
-1 ≥ -5
Let x = -2, y = -2, z = -2
Then xyz = -8
x + y + z - 2 = -8
-8 ≥ -8
Consider x = 2, y = 2, z = 2
Δ = |2 1 1|
|1 2 1|
|1 1 2|
= 2(4 - 1) - 1(2 - 1) + 1(1 - 2)
= 6 - 1 - 1 = 4 ≥ 0
Consider x = -2, y = -2, z = -2
Δ = |-2 1 1|
|1 -2 1|
|1 1 -2|
= -2(4 - 1) - 1(-2 - 1) + 1(1 + 2)
= -6 + 3 + 3 = 0 ≥ 0
The least value of xyz is -8.