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in Linear Inequations by (23.5k points)
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If a2 + b2 + c2 = 1, x2 + y2 + z2 = 1, where a, b, c, x, y, z are positive reals then ax + by + cz is

(a) >

(b) >

(c) <

(d) None of these

1 Answer

+1 vote
by (24.0k points)
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Best answer

(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.

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