(c) xy + yz + zx
Since x > 0, y > 0, z > 0, therefore
(x – y)2 > 0, (y – z)2 > 0, (z – x)2 > 0
⇒ x2 + y2 – 2xy > 0
y2 + z2 – 2yz > 0
z2 + x2 – 2zx > 0
⇒ x2 + y2 > 2xy, y2 + z2 > 2yz, z2 + x2 > 2zx
⇒ x2 + y2 + y2 + z2 + z2 + x2 > 2xy + 2yz + 2zx
⇒ 2(x2 + y2 + z2) > 2(xy + yz + zx)
⇒ x2 + y2 + z2 > xy + yz + zx.