(c) < 1
Applying AM > GM, we have
\(\frac{a+x}{2}≥\sqrt{ax}\) ⇒ a + ≥ 2\(\sqrt{ax}\)
⇒ (a + x)2 > 4ax ⇒ a2 + x2 + 2ax > 4ax
⇒ a2 + x2 > 2ax
Similarly, b2 + y2 > 2by
c2 + z2 > 2cz
∴ a2 + x2 + b2 + y2 + c2 + z2 > 2ax + 2by + 2cz
⇒ (a2 + b2 + c2) + (x2 + y2 + z2) > 2 (ax + by + cz)
⇒ 2 > 2 (ax + by + cz) ⇒ 1 > ax + by + cz
⇒ ax + by + cz < 1.