若x,y,z∈（0,1）且xy+yz+xz=1 求证 y⼀(2-√3x)+z⼀(2-√3y)+

为了记号简便, 换元a = √3·x, b = √3·y, c = √3·z.
则a, b, c ∈ (0,√3), 且ab+bc+ca = 3.
所证不等式变为b/(2-a)+c/(2-b)+a/(2-c) ≥ 3.

由0 < a, b, c < 2, 根据Cauchy不等式有:
(b(2-a)+c(2-b)+a(2-c))(b/(2-a)+c/(2-b)+a/(2-c)) ≥ (a+b+c)².
即b/(2-a)+c/(2-b)+a/(2-c) ≥ (a+b+c)²/(b(2-a)+c(2-b)+a(2-c))
= (a+b+c)²/(2(a+b+c)-(ab+bc+ca))
= (a+b+c)²/(2(a+b+c)-3).

而(a+b+c)² = a²+b²+c²+2ab+2bc+2ca ≥ 3(ab+bc+ca) = 9, 得a+b+c ≥ 3.
故t = 2(a+b+c)-3 > 0.
于是(a+b+c)²/(2(a+b+c)-3) = (t+3)²/(4t) = t/4+9/(4t)+3/2 ≥ 2√(t/4·9/(4t))+3/2 = 3.
即有b/(2-a)+c/(2-b)+a/(2-c) ≥ 3, 证毕.