RMO 2016

Algebra Level 3

Let a , b a,b and c c be 3 distinct positive numbers such that their product is 1. Find the infimium value of

a 3 ( a − b ) ( a − c ) + b 3 ( b − c ) ( b − a ) + c 3 ( c − a ) ( c − b ) . \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-c)(b-a)} + \dfrac{c^3}{(c-a)(c-b)} .


The answer is 3.

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1 solution

Viki Zeta
Oct 18, 2016

Follow up this solution, same question.

Anyway here again:

> a 3 ( a − b ) ( a − c ) + b 3 ( b − a ) ( b − c ) + c 3 ( c − a ) ( c − b ) = a 3 ( a − b ) ( a − c ) − b 3 ( a − b ) ( b − c ) + c 3 ( a − c ) ( b − c ) = a 3 ( b − c ) − b 3 ( a − c ) + c 3 ( a − b ) ( a − b ) ( a − c ) ( b − c ) = a 3 b − a 3 c − b 3 a + b 3 c + c 3 a − c 3 b ( a − b ) ( a − c ) ( b − c ) = a 3 b − b 3 a − a 3 c + b 3 c + c 3 a − c 3 b ) ( a − b ) ( a − c ) ( b − c ) = a b ( a 2 − b 2 ) − c ( a 3 − b 3 ) + c 3 ( a − b ) ( a − b ) ( a − c ) ( b − c ) = a b ( a + b ) ( a − b ) − c ( a − b ) ( a 2 + a b + b 2 ) + c 3 ( a − b ) ( a − b ) ( a − c ) ( b − c ) = ( a − b ) a b ( a + b ) − c ( a 2 + a b + b 2 ) + c 3 ( a − b ) ( a − c ) ( b − c ) = a 2 b + b 2 a − a 2 c − a b c − b 2 c + c 3 ( a − c ) ( b − c ) = a 2 b − a 2 c − b 2 c + c 3 + b 2 a − a b c ( a − c ) ( b − c ) = a 2 ( b − c ) − c ( b 2 − c 2 ) + a b ( b − c ) ( a − c ) ( b − c ) = a 2 ( b − c ) − c ( b + c ) ( b − c ) + a b ( b − c ) ( a − c ) ( b − c ) = a 2 − c ( b + c ) + a b ( a − c ) = a 2 − b c − c 2 + a b ( a − c ) = a 2 − c 2 − b c + a b ( a − c ) = ( a + c ) ( a − c ) + b ( a − c ) ( a − c ) = a + b + c ∴ a 3 ( a − b ) ( a − c ) + b 3 ( b − a ) ( b − c ) + c 3 ( c − a ) ( c − b ) = a + b + c Using AM-GM inequality a + b + c 3 ≥ a b c 3 a b c = 1 ⟹ a + b + c 3 ≥ 1 ⟹ a + b + c ≥ 3 ∴ a 3 ( a − b ) ( a − c ) + b 3 ( b − a ) ( b − c ) + c 3 ( c − a ) ( c − b ) = a + b + c ≥ 3 \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)}\\ = \dfrac{a^3}{(a-b)(a-c)} - \dfrac{b^3}{(a-b)(b-c)} + \dfrac{c^3}{(a-c)(b-c)} \\ = \dfrac{a^3(b-c) - b^3(a-c) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^3b - a^3c - b^3a + b^3c + c^3a - c^3b}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^3b - b^3a - a^3c + b^3c+c^3a - c^3b)}{(a-b)(a-c)(b-c)}\\ = \dfrac{ab(a^2 - b^2) -c(a^3 - b^3) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = \dfrac{ab(a+b)(a-b) - c (a-b)(a^2 + ab +b^2) + c^3(a-b)}{(a-b)(a-c)(b-c)} \\ = (a-b)\dfrac{ab(a+b) - c(a^2 + ab + b^2) + c^3}{(a-b)(a-c)(b-c)} \\ = \dfrac{a^2b + b^2a - a^2c - abc - b^2c + c^3}{(a-c)(b-c)} \\ = \dfrac{a^2b - a^2c - b^2c + c^3 + b^2a - abc}{(a-c)(b-c)}\\ \\ = \dfrac{a^2(b-c) -c(b^2 - c^2) + ab(b-c)}{(a-c)(b-c)}\\ = \dfrac{a^2(b-c) - c(b+c)(b-c) + ab(b-c)}{(a-c)(b-c)}\\ = \dfrac{a^2 - c(b+c) + ab}{(a-c)} \\ = \dfrac{a^2 - bc - c^2 + ab}{(a-c)} \\ = \dfrac{a^2 - c^2 - bc + ab}{(a-c)} \\ = \dfrac{(a+c)(a-c) + b(a-c)}{(a-c)} \\ = a+b+c \\ \boxed{\therefore \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)} = a+b+c} \\ \text{Using AM-GM inequality} \\ \dfrac{a+b+c}{3} \ge \sqrt[3]{abc} \\ abc = 1 \\ \implies \dfrac{a+b+c}{3} \ge 1 \\ \implies a+b+c \ge 3 \\ \boxed{\therefore \dfrac{a^3}{(a-b)(a-c)} + \dfrac{b^3}{(b-a)(b-c)} + \dfrac{c^3}{(c-a)(c-b)} = a+b+c \ge 3}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

As with any inequality, to find the minimium, we not only have to show that it is a lower bound, but we also have to show that it can indeed be achieved.

Particularly in this problem, because we require distinct values, hence we cannot achieve 3. In this problem, there is no minimum, though there is an infimium.

Calvin Lin Staff - 4 years, 7 months ago

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Yeah, the same case in my problem too. Thanks.

Viki Zeta - 4 years, 7 months ago

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