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Saudi Booklet

Algebra Difficulty 4.5 AIME Prove it Saudi Arabia

Let a,b,ca, b, c be nonnegative numbers such that a≥b≥ca \ge b \ge c. Prove that
a3+b3+c3−3abc≥92(a−b)(b2−c2). a^3 + b^3 + c^3 - 3abc \ge \frac{9}{2}(a-b)(b^2-c^2).

Solution

Write the inequality as
(a+b+c)(a2+b2+c2−ab−bc−ca)≥92(a−b)(b−c)(b+c). (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \ge \frac{9}{2}(a - b)(b - c)(b + c).
Since a+b+c≥32(b+c)a + b + c \ge \frac{3}{2}(b + c), it suffices to show that
a2+b2+c2−ab−bc−ca≥3(a−b)(b−c). a^2 + b^2 + c^2 - ab - bc - ca \ge 3(a - b)(b - c).
This is equivalent to the obvious inequality (a−2b+c)2≥0(a - 2b + c)^2 \ge 0. The equality holds for a=b=ca = b = c. □\square

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