1/2 = (x+y+z)/3 ≥
3√xyz eqv. at x=y=z=1/2 | AM ≥ GM
=>
3√[1/(xyz)] ≥ 2 eqv. at x=y=z=1/2
=> (1/x+1/y+1/z)/3 ≥
3√[1/(xyz)] ≥ 2 eqv. at x=y=z=1/2, (1/x)=(1/y)=(1/z)=2 | AM ≥ GM
=> [(x+1/x)+(y+1/y)+(z+1/z)]/3 = (x+y+z)/3+(1/x+1/y+1/z)/3 ≥ 1/2+2 = 5/2
eqv. at x=y=z=1/2, (1/x)=(1/y)=(1/z)=2, (x+1/x)=(y+1/y)=(z+1/z)=5/2
i.e.
3√[(x+1/x)(y+1/y)(z+1/z)] ≥ 3/[1/(x+1/x)+1/(y+1/y)+1/(z+1/z)]
eqv. at (x+1/x)=(y+1/y)=(z+1/z)=5/2 | GM ≥ HM
≥ 3/{[1/(5/2)+1/(5/2)+1/(5/2)]} |(x+1/x=y+1/y=z+1/z=5/2)
= 5/2
∴ (x+1/x)(y+1/y)(z+1/z) ≥ (5/2)
3 = 125/8