IMO 1975 Q1 #
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Let x₁, x₂, ... , xₙ
and y₁, y₂, ... , yₙ
be two sequences of real numbers, such that
x₁ ≥ x₂ ≥ ... ≥ xₙ
and y₁ ≥ y₂ ≥ ... ≥ yₙ
. Prove that if z₁, z₂, ... , zₙ
is any permutation
of y₁, y₂, ... , yₙ
, then ∑ (xᵢ - yᵢ)^2 ≤ ∑ (xᵢ - zᵢ)^2
Solution #
Firstly, we expand the squares withing both sums and distribute into separate finite sums. Then,
noting that ∑ yᵢ ^ 2 = ∑ zᵢ ^ 2
, it remains to prove that ∑ xᵢ * zᵢ ≤ ∑ xᵢ * yᵢ
, which is true
by the Rearrangement Inequality
theorem
imo1975_q1
(n : ℕ)
(σ : equiv.perm ℕ)
(hσ : {x : ℕ | ⇑σ x ≠ x} ⊆ ↑(finset.Icc 1 n))
(x y : ℕ → ℝ)
(hx : antitone_on x ↑(finset.Icc 1 n))
(hy : antitone_on y ↑(finset.Icc 1 n)) :