We know that,
AM≥HM
Considering (x+y),(y+z) and (x+z) as three numbers, we get
(x+y)+(y+z)+(x+z)3≥31x+y+1y+z+1x+z
⇒2(x+y+z)(1x+y+1y+z+1x+z)≥9
⇒1+zx+y+1+xy+z+1+yz+x≥92
⇒x1−x+y1−y+z1−z≥32
⇒2x1−x+2y1−y+2z1−z≥3
Hence, the least value of the given expression is 3.