07/06/10 20:28:38
>>51 下
[A.425]
Let n≧2 and let a_1,a_2,…,a_n, x_1,x_2,…,x_n be positive real numbers such that a_1+a_2+ … +a_n =A, x_1+x_2+ … + x_n =X.
Prove that
2Σ[1≦i<j≦n] x_i･x_j ≦ {(n-2)/(n-1)}X^2 + Σ[i=1,n] {a_i/(A-a_i)}(x_i)^2.
(略解)
コーシーより
Σ[i=1,n] {A/(A-a_i)} (x_i)^2 ≧{Σ[k=1,n] x_k}^2 /{Σ[j=1,n] (A-a_j)/A } = {1/(n-1)}X^2,
よって
(右辺) = {(n-2)/(n-1)}X^2 + Σ[i=1,n] {A/(A-a_i)}(x_i)^2 - Σ[i=1,n] (x_i)^2
≧ {(n-2)/(n-1)}X^2 + {1/(n-1)}X^2 - Σ[i=1,n] (x_i)^2
= X^2 - Σ[i=1,n] (x_i)^2
= 2Σ[1≦i<j≦n] x_i･x_j
= (左辺).
[B.3997] >>58
[B.4000]
Find the smallest possible value of x^2 +y^2, given that x and y are real numbers, x≠0 and xy(x^2 -y^2) = x^2 +y^2.
(略解)
(1/4)(x^2 +y^2)^2 = (1/4){(2xy)^2 + (x^2 -y^2)^2} ≧ (1/2)(2xy)(x^2 -y^2) = xy(x^2 -y^2)
と与式から
x^2 +y^2 ≧ 4,
等号成立は 2xy = x^2 -y^2,
(x,y) = (±√(2+√2), ±√(2-√2)), (±√(2-√2), 干√(2+√2)) 〔複号同順〕
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