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Algebra Difficulty 7.2 National Olympiad, round 2 Prove it Turkey

Determine the smallest value of (a+5)2+(b−2)2+(c−9)2(a+5)^2 + (b-2)^2 + (c-9)^2 for all real numbers a,b,ca, b, c satisfying a2+b2+c2−ab−bc−ca=3a^2 + b^2 + c^2 - ab - bc - ca = 3.

Solution

Note that (a+5)2+(b−2)2+(c−9)2≥72(a+5)^2 + (b-2)^2 + (c-9)^2 \ge 72 since
(a+5)2+(b−2)2+(c−9)2+4⋅3−84=(a+5)2+(b−2)2+(c−9)2+4(a2+b2+c2−ab−bc−ca)−84=(a−2b+3)2+(b−2c+4)2+(c−2a−1)2≥0 (a+5)^2 + (b-2)^2 + (c-9)^2 + 4 \cdot 3 - 84 = (a+5)^2 + (b-2)^2 + (c-9)^2 + 4(a^2+b^2+c^2-ab-bc-ca) - 84 \\ = (a - 2b + 3)^2 + (b - 2c + 4)^2 + (c - 2a - 1)^2 \ge 0
The equality is held at (a,b,c)=(1,2,3)(a, b, c) = (1, 2, 3).

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