Algebraic inequality 1 Let x, y, Z ∈ R +, prove: X √ [x / (1 + YZ)] + y √ [y / (1 + ZX)] + Z √ [Z / (1 + XY)] ≥ 3 / √ (1 + XYZ)

Algebraic inequality 1 Let x, y, Z ∈ R +, prove: X √ [x / (1 + YZ)] + y √ [y / (1 + ZX)] + Z √ [Z / (1 + XY)] ≥ 3 / √ (1 + XYZ)

If x = y = Z and tends to 0, then the left end tends to 0 and the right end tends to 3
If we add the condition x + y + Z = 3, it is not difficult to prove
Firstly, the inequality can be reduced to X & # 178; / √ (x + XYZ) + Y & # 178; / √ (y + XYZ) + Z & # 178; / √ (Z + XYZ) ≥ 3 / √ (1 + XYZ)
From the Cauchy inequality, we get that:
(√(x+xyz)+√(y+xyz)+√(z+xyz))(x²/√(x+xyz)+y²/√(y+xyz)+z²/√(z+xyz)) ≥ (x+y+z)² = 9.
It is also proved by Cauchy inequality that:
9(1+xyz) = (1+1+1)((x+xyz)+(y+xyz)+(z+xyz)) ≥ (√(x+xyz)+√(y+xyz)+√(z+xyz))²,
That is, 3 √ (1 + XYZ) ≥ √ (x + XYZ) + √ (y + XYZ) + √ (Z + XYZ)
So x & # 178; / √ (x + XYZ) + Y & # 178; / √ (y + XYZ) + Z & # 178; / √ (Z + XYZ)
≥ 9/(√(x+xyz)+√(y+xyz)+√(z+xyz))
≥ 3/√(1+xyz),
That is the proof