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Algebra Difficulty 5.7 AIME, harder Prove it China

Let n≥3n \ge 3 be integer. Suppose that α,β,γ∈(0,1)\alpha, \beta, \gamma \in (0, 1) and ak,bk,ck≥0a_k, b_k, c_k \ge 0 (k=1,2,…,n)(k = 1, 2, \dots, n) satisfy ∑k=1n(k+α)ak≤α\sum_{k=1}^n (k+\alpha)a_k \le \alpha, ∑k=1n(k+β)bk≤β\sum_{k=1}^n (k+\beta)b_k \le \beta and ∑k=1n(k+γ)ck≤γ\sum_{k=1}^n (k+\gamma)c_k \le \gamma. Find the minimum of λ\lambda such that ∑k=1n(k+λ)akbkck≤λ\sum_{k=1}^n (k+\lambda)a_k b_k c_k \le \lambda.

Solution

Let a1=α1+αa_1 = \frac{\alpha}{1+\alpha}, b1=β1+βb_1 = \frac{\beta}{1+\beta}, c1=γ1+γc_1 = \frac{\gamma}{1+\gamma}, ai,bi,ci=0a_i, b_i, c_i = 0 (i=2,3,…,n)(i = 2, 3, \dots, n). We see that all conditions are satisfied. So, we must have
(1+λ)α1+α⋅β1+β⋅γ1+γ≤λ, (1 + \lambda) \frac{\alpha}{1 + \alpha} \cdot \frac{\beta}{1 + \beta} \cdot \frac{\gamma}{1 + \gamma} \le \lambda,
that is,
λ≥αβγ(1+α)(1+β)(1+γ)−αβγ. \lambda \ge \frac{\alpha\beta\gamma}{(1+\alpha)(1+\beta)(1+\gamma) - \alpha\beta\gamma}.

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