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x + y + z + 1/2x yz >= 4 when x^2 +y^2 + z^2 = 2(x y + yz + z x)

Source: 2016 Saudi Arabia Pre-TST Level 4 1.1

September 13, 2020
algebrainequalities

Problem Statement

Let x,y,zx, y, z be positive real numbers satisfy the condition x2+y2+z2=2(xy+yz+zx)x^2 +y^2 + z^2 = 2(x y + yz + z x). Prove that x+y+z+12xyz≥4x + y + z + \frac{1}{2x yz} \ge 4
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