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Problem 1261

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Algebra Difficulty 5.3 Prove it Indian National Mathematical Olympiad · India

Determine the least positive value taken by the expression a3+b3+c33abca^{3}+b^{3}+c^{3}-3 a b c as a,b,ca, b, c vary over all positive integers. Find also all triples (a,b,c)(a, b, c) for which this least value is attained.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

We observe that
Q=a3+b3+c33abc=12(a+b+c)((ab)2+(bc)2+(ca)2) Q = a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2}(a + b + c)\left((a - b)^{2} + (b - c)^{2} + (c - a)^{2}\right)
Since we are looking for the least positive value taken by QQ, it follows that a,b,ca, b, c are not all equal. Thus a+b+c1+1+2=4a + b + c \geq 1 + 1 + 2 = 4 and (ab)2+(bc)2+(ca)21+1+0=2(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 1 + 1 + 0 = 2. Thus we see that Q4Q \geq 4. Taking a=1a = 1, b=1b = 1 and c=2c = 2, we get Q=4Q = 4. Therefore the least value of QQ is 44 and this is achieved only by a+b+c=4a + b + c = 4 and (ab)2+(bc)2+(ca)2=2(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 2. The triples for which Q=4Q = 4 are therefore given by
(a,b,c)=(1,1,2), (1,2,1), (2,1,1) (a, b, c) = (1, 1, 2),\ (1, 2, 1),\ (2, 1, 1)

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.