IMO 2004 SL A5

Origin: THA | Category: Algebra

Problem

Let a, b, c > 0 and ab + bc + ca = 1. Prove the inequality  a + 6b +  b + 6c +  c + 6a \leq abc.

Solution

By the general mean inequality (M1 \leqM3), the LHS of the inequality to be proved does not exceed E = 3\sqrt  a + 1 b + 1 c + 6(a + b + c). From ab + bc + ca = 1 we obtain that 3abc(a + b + c) = 3(ab \cdot ac + ab \cdot bc + ac \cdot bc) \leq(ab + ac + bc)2 = 1; hence 6(a + b + c) \leq abc. Since a + 1 b + 1 c = ab+bc+ca abc

abc, it follows that E \leq 3\sqrt  abc \leq abc, where the last inequality follows from the AM–GM inequality 1 = ab+bc+ ca \geq3 3 (abc)2, i.e., abc \leq1/(3 \sqrt 3). The desired inequality now follows. Equality holds if and only if a = b = c = 1/ \sqrt 3.