The value of the greatest integer k satisfying the inequation 2n+4 + 12 ≥ k(n + 4) for all n ∈ N is:

The inequality |x| ≥ 5 implies that:

Consider positive numbers a1 ≥ a2 ≥ a3 and b1 ≥ b2 ≥ b3. If Σ ai = 12 and Σ bi = 18, then the minimum possible value of a1b3 + a2b2 + a3b1 according to Chebyshev's sum inequality is:

If |x - 4| + |x - 2| < 4, find the solution set for x.