分类：教育/科学-理工学科-数学
问题：求极值设x,y,z为正数，满足:x^2+y^2+z^2=1，求x 
设x,y,z为正数，满足:x^2+y^2+z^2=1，求x/(1-x^2)+y/(1-y^2)+z/(1-z^2) 的极值.

回答：证明 容易猜到当x=y=z=√(1/3)时，
x/(1-x^2)+y/(1-y^2)+z/(1-z^2)取得最小值(3√3)/2.
据柯西不等式得:
[x/(1-x^2)+y/(1-y^2)+z/(1-z^2)]*[x^3*(1-x^2)+y^3*(1-y^2)+z^3*(1-z^2)]
≥(x^2+y^2+z^2)^2=1.
故只需证明
x^3*(1-x^2)+y^3*(1-y^2)+z^3*(1-z^2)≤2/(3√3)   (1)
(1)<==> 2/(3√3)+x^5+y^5+z^5≥x^3+y^3+z^3       (2)
由均值不等式得:
2x^2/(3√3)+x^5≥3*{[x^2/(3√3)]^2*x^5}^(1/3)=x^3;  3-1)
2x^2/(3√3)+z^5≥3*{[z^2/(3√3)]^2*z^5}^(1/3)=z^3; (3-2)
2y^2/(3√3)+y^5≥3*{[y^2/(3√3)]^2*y^5}^(1/3)=y^3  (3-3) 
(3-1)+(3-2)+(3-3)即得(2)式。

下面给出第二种证法:
求x/(1-x^2)+y/(1-y^2)+z/(1-z^2) 的最小值.只需证:
x/(1-x^2)+y/(1-y^2)+z/(1-z^2)≥(3√3)/2.   (1)
因为x,y,z为正数，所以
x/(1-x^2)-(3√3)x^2/2=x(√3*x-1)^2*(√3*x+2)/[2(1-x^2)]≥0
故得:
x/(1-x^2)≥[(3√3)/2]*x^2  (2)
z/(1-z^2)≥[(3√3)/2]*z^2  (3)
t/(1-y^2)≥[(3√3)/2]*y^2  (4)
(2)+(3)+(4)得:
x/(1-x^2)+y/(1-y^2)+z/(1-z^2)≥[(3√3)/2]*(x^2+y^2+z^2)=(3√3)/2.