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Romanian Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Romania

Let aa, bb, cc be integers larger than 11. Prove that
a(a−1)+b(b−1)+c(c−1)≤(a+b+c−4)(a+b+c−5)+4. a(a-1) + b(b-1) + c(c-1) \leq (a + b + c - 4)(a + b + c - 5) + 4.

Solution

The inequality is equivalent to
a2+b2+c2−a−b−c≤(a+b+c)2−9(a+b+c)+24, a^2 + b^2 + c^2 - a - b - c \leq (a + b + c)^2 - 9(a + b + c) + 24,
or
0≤ab+bc+ca−4(a+b+c)+12. 0 \leq ab + bc + ca - 4(a + b + c) + 12.
Since ab−2a−2b+4=(a−2)(b−2)ab - 2a - 2b + 4 = (a - 2)(b - 2), summing together with the similar relations implies (ab−2a−2b+4)+(bc−2b−2c+4)+(ca−2c−2a+4)=(a−2)(b−2)+(b−2)(c−2)+(c−2)(a−2)≥0(ab - 2a - 2b + 4) + (bc - 2b - 2c + 4) + (ca - 2c - 2a + 4) = (a - 2)(b - 2) + (b - 2)(c - 2) + (c - 2)(a - 2) \geq 0, since aa, bb, c≥2c \geq 2.

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