Olympiad Maths Prep

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, 2010

Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Nonnegative real numbers aa, bb, cc satisfy the equation a+b+c=1a + b + c = 1. Prove the inequality
(1−a)2+(1−b)2+(1−c)2≥6abc. (1 - a)^2 + (1 - b)^2 + (1 - c)^2 \geq 6\sqrt{abc}.

Can the case of equality occur?

Solution

Removing the brackets we will have (1−a)2+(1−b)2+(1−c)2=3+a2+b2+c2−2(a+b+c)+3=a2+b2+c2+a+b+c≥6abc(1 - a)^2 + (1 - b)^2 + (1 - c)^2 = 3 + a^2 + b^2 + c^2 - 2(a + b + c) + 3 = a^2 + b^2 + c^2 + a + b + c \geq 6\sqrt{abc}, equality is achieved when a=b=c=a2=b2=c2a = b = c = a^2 = b^2 = c^2, which is impossible, because a+b+c=1a + b + c = 1.

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