Maths Olympiad Prep

Track / Stage 7 / 132 of 300 #1532 of 1964

Problem 1532

National Olympiad second round; IMO P1/P4
Algebra Difficulty 7.3 Find the answer

n≥3n\geq 3 is a integer. α,β,γ∈(0,1)\alpha,\beta,\gamma \in (0,1). For every ak,bk,ck≥0(k=1,2,…,n)a_k,b_k,c_k\geq0(k=1,2,\dotsc,n) with ∑k=1n(k+α)ak≤α,∑k=1n(k+β)bk≤β,∑k=1n(k+γ)ck≤γ\sum\limits_{k=1}^n(k+\alpha)a_k\leq \alpha, \sum\limits_{k=1}^n(k+\beta)b_k\leq \beta, \sum\limits_{k=1}^n(k+\gamma)c_k\leq \gamma, we always have ∑k=1n(k+λ)akbkck≤λ\sum\limits_{k=1}^n(k+\lambda)a_kb_kc_k\leq \lambda.
Find the minimum of λ\lambda

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Official solution

1. Given Conditions and Initial Setup:
We are given that n≥3 n \geq 3 is an integer and α,β,γ∈(0,1) \alpha, \beta, \gamma \in (0,1) . For every ak,bk,ck≥0 a_k, b_k, c_k \geq 0 (for k=1,2,…,n k = 1, 2, \dotsc, n ) with the constraints:
∑k=1n(k+α)ak≤α,∑k=1n(k+β)bk≤β,∑k=1n(k+γ)ck≤γ, \sum_{k=1}^n (k+\alpha)a_k \leq \alpha, \quad \sum_{k=1}^n (k+\beta)b_k \leq \beta, \quad \sum_{k=1}^n (k+\gamma)c_k \leq \gamma,
we need to show that:
∑k=1n(k+λ)akbkck≤λ. \sum_{k=1}^n (k+\lambda)a_k b_k c_k \leq \lambda.
Our goal is to find the minimum value of λ\lambda.

2. Initial Inequality Analysis:
Since ∑abc≤∑(k+α)a∑(k+β)b∑(k+γ)c≤αβγ<1\sum abc \leq \sum (k+\alpha)a \sum (k+\beta)b \sum (k+\gamma)c \leq \alpha \beta \gamma < 1, we can infer that:
λ≥∑kabc1−∑abc. \lambda \geq \frac{\sum k a b c}{1 - \sum a b c}.
We need to find the maximum of:
I=∑kabc1−∑abc. I = \frac{\sum k a b c}{1 - \sum a b c}.

3. Expression Simplification:
We express II in terms of α,β,γ\alpha, \beta, \gamma:
I=αβγ∑kabcαβγ−αβγ∑abc. I = \alpha \beta \gamma \frac{\sum k a b c}{\alpha \beta \gamma - \alpha \beta \gamma \sum a b c}.
Using the given constraints, we have:
I≤αβγ∑kabc∑(k+α)a∑(k+β)b∑(k+γ)c−αβγ∑abc. I \leq \frac{\alpha \beta \gamma \sum k a b c}{\sum (k+\alpha)a \sum (k+\beta)b \sum (k+\gamma)c - \alpha \beta \gamma \sum a b c}.
This can be further simplified to:
I≤αβγ∑kabc∑[(k+α)(k+β)(k+γ)−αβγ]abc. I \leq \frac{\alpha \beta \gamma \sum k a b c}{\sum [(k+\alpha)(k+\beta)(k+\gamma) - \alpha \beta \gamma] a b c}.

4. Maximization and Final Bound:
Using the fact that x+ya+b≤max⁡(xa,yb)\frac{x+y}{a+b} \leq \max\left(\frac{x}{a}, \frac{y}{b}\right), we get:
I≤αβγmax⁡1≤k≤nk(k+α)(k+β)(k+γ)−αβγ. I \leq \alpha \beta \gamma \max_{1 \leq k \leq n} \frac{k}{(k+\alpha)(k+\beta)(k+\gamma) - \alpha \beta \gamma}.
Therefore, we have:
I≤αβγ(1+α)(1+β)(1+γ)−αβγ. I \leq \frac{\alpha \beta \gamma}{(1+\alpha)(1+\beta)(1+\gamma) - \alpha \beta \gamma}.

5. Attainability:
This bound is attainable by setting:
a1=α1+α,b1=β1+β,c1=γ1+γ, a_1 = \frac{\alpha}{1+\alpha}, \quad b_1 = \frac{\beta}{1+\beta}, \quad c_1 = \frac{\gamma}{1+\gamma},
and setting all other ak,bk,cka_k, b_k, c_k to zero.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.