Q.94. Algebraic Inequality.

Question 94.
Suppose a,b,c are real numbers such that a^2 * b^2 + b^2 * c^2 + c^2 * a^2 = k, where k is a constant,
then prove that - k/2 ≤ abc(a + b + c) ≤ k.

Answer 94.
(ab + bc + ca)^2 ≥ 0
=> a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) ≥ 0
=> k + 2abc(a + b + c) ≥ 0
=> - k/2 ≤ abc(a + b + c) ... ( 1 )

(ab - bc)^2 + (bc - ca)^2 + (ca - ab)^2 ≥ 0
=> 2(a^2b^2 + b^2c^2 + c^2a^2) - 2abc(a + b + c) ≥ 0
=> k - abc(a + b + c) ≥ 0
=> abc(a + b + c) ≤ k ... ( 2 )

From equations ( 1 ) and ( 2 ),
- k/2 ≤ abc(a + b + c) ≤ k.

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