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A Generalisation on an RMO Problem

\[\dfrac {a^k}{(a-b)(a-c)} + \dfrac {b^k}{(b-a)(b-c)} + \dfrac {c^k}{(c-a)(c-b)} > \dfrac {k(k-1)}{2}\]

If \(a\), \(b\) and \(c\) are distinct positive reals such that \(abc = 1\), and \(k \geq 3\) is a positive integer, show that the inequality above holds.

Note: The RMO problem had the case \(k=3\).

Note by Sharky Kesa
3 months ago

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Sharky, you've got the right idea. There is a much easier way to show that identity

Claim: \[ a^m (b-c) + b^m(c-a) + c^m (a-b) = (a^2b-a^2c+b^2c-b^2a+c^2a-c^2b) \sum_{i + j + k = m-2} a^i b^j c^k \]

Proof: Let's find the coefficient of the term \( a^p b^q c^r, \) where \( p+q+r = m+1, \) on the RHS.

  • If \( 2 \leq p, q, r \), then the coefficient would be 0, because we can obtain it via multiplying through all 6 terms in the first factor.
  • If \( \{p,q,r\} = \{0,0, m+1\} \), then the coefficient would be 0 because we are multiplying by cross terms in the first factor.
  • If \( \{p,q,r\} = \{ 0, 1, m \} \), show that it has the desired coefficient.
  • If \( \{ p, q, r \} = \{ 1, 1, m-1 \} \), show that it has the desired coefficient of 0.

Sometimes if you know what you want, it's best to prove it directly instead of trying to explain how you found it out. Esp if the way that you first found it provides no additional insight into the process. Calvin Lin Staff · 2 months, 3 weeks ago

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Anyone got a better solution than this?

Firstly, we add the fractions together (using the fact \(b-a=-(a-b)\), etc. to get

\[\begin{align} \dfrac {a^k}{(a-b)(a-c)} + \dfrac {b^k}{(b-a)(b-c)} + \dfrac {c^k}{(c-a)(c-b)} &= \dfrac{a^k (b-c) - b^k (a-c) + c^k (a-b)}{(a-b)(a-c)(b-c)}\\ &= \dfrac {a^k b - ab^k - a^k c + b^k c + c^k (a-b)}{(a-b)(a-c)(b-c)}\\ &= \dfrac {ab (a-b)(a^{k-2} + \ldots + b^{k-2}) - c (a-b)(a^{k-1} + \ldots + b^{k-1}) + c^k (a-b)}{(a-b)(a-c)(b-c)}\\ &= \dfrac {a^{k-1} b + \ldots + ab^{k-1} - a^{k-1} c + \ldots + b^{k-1} c + c^k}{(a-c)(b-c)}\\ &= \dfrac {a^{k-1} b - a^{k-1} c + a^{k-2} b^2 - a^{k-2} bc + \ldots + ab^{k-1} - ab^{k-2} c - b^{k-1} c + c^k}{(a-c)(b-c)}\\ &= \dfrac {(b-c)(a^{k-1} + a^{k-2}b + \ldots + ab^{k-2}) - c(b-c)(b^{k-2} + \ldots + c^{k-2})}{(a-c)(b-c)}\\ &= \dfrac {a^{k-1} + \ldots + ab^{k-2} - b^{k-2}c - \ldots - c^{k-1}}{a-c}\\ &= \dfrac {a^{k-1} - c^{k-1} + a^{k-2} b - b c^{k-2} + \ldots + ab^{k-2} - b^{k-2} c}{a-c}\\ &= \dfrac {(a-c)(a^{k-2} + \ldots + c^{k-2}) + b(a-c)(a^{k-3} + \ldots + c^{k-3}) + \ldots + b^{k-2} (a-c)}{a-c}\\ &= a^{k-2} + \ldots + c^{k-2} + a^{k-3} b + \ldots + b c^{k-3} + \ldots + b^{k-2} \end{align}\]

Notice that this expression goes through every single possible term composed of variables \(a, b, c\) and degree \(k-2\). Thus, by Supermarket Principle, there are \(\binom{k}{2}\) terms here. Notice that the product of all the terms will have equal degrees for each of \(a\), \(b\) and \(c\) by symmetry, so their product will be just 1 (using \(abc = 1\)). Also, we must have \(a\), \(b\) and \(c\) distinct for the original expression to have a defined value. Thus, by AM-GM, we have

\[\begin{align} \dfrac {a^{k-2} + \ldots + c^{k-2} + a^{k-3} b + \ldots + b c^{k-3} + \ldots + b^{k-2}}{\dbinom{k}{2}} &> 1\\ a^{k-2} + \ldots + c^{k-2} + a^{k-3} b + \ldots + b c^{k-3} + \ldots + b^{k-2} &> \dbinom {k}{2} = \dfrac {k(k-1)}{2} \end{align}\]

Thus, proven. Sharky Kesa · 2 months, 3 weeks ago

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@Svatejas Shivakumar To be clear, your approach works for general \(n\) through a simple generalization. Calvin Lin Staff · 2 months, 4 weeks ago

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Comment deleted 3 months ago

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@Svatejas Shivakumar The claim of "One can prove by induction" is not obvious, in part since you didn't use the induction hypothesis to prove \( k = 4 \).

Can you elaborate on the induction approach?

Note: The approach that I'm thinking of simply uses what you have, without any inductive step. It essentially boils down to finding that "4th factor" and showing that it is greater than k(k-1)/2. Calvin Lin Staff · 3 months ago

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