If x,y,z∈R+ then xyx+y+yzy+z+xzx+z is always
≤12(x+y+z)
From A.M. ≥ G.M.≥ H.M.
We know that A.M. ≥ H.M. and G.M. ≥ H.M.
x+y2≥2xyx+y
⇒ xyx+y≤14(x+y)⋯(1)
y+z2≥2yzy+z
⇒ yzy+z≤14(y+z)⋯(2)
Similarly, xzx+z≤14(x+z)⋯(3)
Adding (1), (2) and (3),
xyx+y+xyx+y+xzx+z≤14(2x+2y+2z)
⇒xyx+y+xyx+y+xzx+z≤12(x+y+z)
Hence, Option b. is true