A Generalisation on an RMO Problem

ak(a−b)(a−c)+bk(b−a)(b−c)+ck(c−a)(c−b)>k(k−1)2\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 aa, bb and cc are distinct positive reals such that abc=1abc = 1, and k≥3k \geq 3 is a positive integer, show that the inequality above holds.

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

#Algebra

Note by Sharky Kesa
4 years, 8 months ago

No vote yet
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Comments

Anyone got a better solution than this?

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

ak(a−b)(a−c)+bk(b−a)(b−c)+ck(c−a)(c−b)=ak(b−c)−bk(a−c)+ck(a−b)(a−b)(a−c)(b−c)=akb−abk−akc+bkc+ck(a−b)(a−b)(a−c)(b−c)=ab(a−b)(ak−2+…+bk−2)−c(a−b)(ak−1+…+bk−1)+ck(a−b)(a−b)(a−c)(b−c)=ak−1b+…+abk−1−ak−1c+…+bk−1c+ck(a−c)(b−c)=ak−1b−ak−1c+ak−2b2−ak−2bc+…+abk−1−abk−2c−bk−1c+ck(a−c)(b−c)=(b−c)(ak−1+ak−2b+…+abk−2)−c(b−c)(bk−2+…+ck−2)(a−c)(b−c)=ak−1+…+abk−2−bk−2c−…−ck−1a−c=ak−1−ck−1+ak−2b−bck−2+…+abk−2−bk−2ca−c=(a−c)(ak−2+…+ck−2)+b(a−c)(ak−3+…+ck−3)+…+bk−2(a−c)a−c=ak−2+…+ck−2+ak−3b+…+bck−3+…+bk−2\begin{aligned} \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{aligned}

Notice that this expression goes through every single possible term composed of variables a,b,ca, b, c and degree k−2k-2. Thus, by Supermarket Principle, there are (k2)\binom{k}{2} terms here. Notice that the product of all the terms will have equal degrees for each of aa, bb and cc by symmetry, so their product will be just 1 (using abc=1abc = 1). Also, we must have aa, bb and cc distinct for the original expression to have a defined value. Thus, by AM-GM, we have

ak−2+…+ck−2+ak−3b+…+bck−3+…+bk−2(k2)>1ak−2+…+ck−2+ak−3b+…+bck−3+…+bk−2>(k2)=k(k−1)2\begin{aligned} \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{aligned}

Thus, proven.

Sharky Kesa - 4 years, 7 months ago

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

Claim: am(b−c)+bm(c−a)+cm(a−b)=(a2b−a2c+b2c−b2a+c2a−c2b)∑i+j+k=m−2aibjck 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 apbqcr, a^p b^q c^r, where p+q+r=m+1, p+q+r = m+1, on the RHS.

  • If 2≤p,q,r 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} \{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} \{p,q,r\} = \{ 0, 1, m \} , show that it has the desired coefficient.
  • If {p,q,r}={1,1,m−1} \{ 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 - 4 years, 7 months ago

@Svatejas Shivakumar To be clear, your approach works for general nn through a simple generalization.

Calvin Lin Staff - 4 years, 7 months ago
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