Question

数学不等式题目求解8

x,y,z> -1,且属于R，求证(1+x^2)/(1+y+z^2)＋(1+y^2)/(1+z+x^2)＋(1+z^2)/(1+x+y^2)>=2

Answer 1

以sum记对三个变量取循环和。将两边同时乘以

S = sum (1+x^2)(1+y+z^2)

则由柯西不等式：

S*左端 >= (sum (1+x^2))^2

另一方面，

y <= (1+y^2)/2 => (1+x^2)(1+y+z^2) <= (1+x^2)(1+z^2) + (1+x^2)(1+y^2)/2

于是

S <= sum (1+x^2)(1+z^2) + (1+x^2)(1+y^2)/2
= (3/2) sum (1+x^2)(1+y^2)
<= (1/2) (sum (1+x^2))^2

从而

(1/2) (sum (1+x^2))^2*左端 >= S*左端 >= (sum (1+x^2))^2

化简即得：左端>=2

Answer 2

x,y,z是轮换对称的，当三者相等时，通常取到最值。
具体的用柯西做吧，其实我是选修柯西的，掌握的不是横好

Answer 3

以sum记对三个变量取循环和。将两边同时乘以

S = sum (1+x^2)(1+y+z^2)

则由柯西不等式：

S*左端 >= (sum (1+x^2))^2

另一方面，

y <= (1+y^2)/2 => (1+x^2)(1+y+z^2) <= (1+x^2)(1+z^2) + (1+x^2)(1+y^2)/2

于是

S <= sum (1+x^2)(1+z^2) + (1+x^2)(1+y^2)/2
= (3/2) sum (1+x^2)(1+y^2)
<= (1/2) (sum (1+x^2))^2

(1/2) (sum (1+x^2))^2*左端 >= S*左端 >= (sum (1+x^2))^2

左端>=2