Since A.M≥G.M
3x+y+z≥(xyz)31
x+y+z≥3(xyz)31
Now, ∣x111y111z∣=xyz−(x+y+z)+2
Now determinant is non-negative
⇒xyz−(x+y+z)+2≥0
⇒xyz−(3)(xyz)31+2≥0
Let xyz=t3
So, t3−3t+2≥0
⇒(t+2)(t2−2t+1)≥0⇒(t+2)(t−1)2≥0
⇒t≥−2
⇒t3≥−8
⇒ Least value of xyz=−8
The least value of the product xyz (such that x,yandz are positive real numbers) for which the determinant ∣x111y111z∣ is non-negative is
Held on 10 Apr 2015 · Verified 6 Jul 2026.
−1
−162
−8
−22
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