x,y,z是正实数,x^2+y^2+z^2=1,求证：x/(1-x^2)+y/(1-y^2)+z/(1-z^2)≥2分之3倍根号3. 请高手解答.谢谢!

题目

x,y,z是正实数,x^2+y^2+z^2=1,求证：x/(1-x^2)+y/(1-y^2)+z/(1-z^2)≥2分之3倍根号3. 请高手解答.谢谢!

答案

显然 0< x^2<1
我们考虑 2x^2*(1-x^2)^2=2x^2*(1-x^2)(1-x^)<=((2x^2+1-x^2+1-x^2)/3)^3=8/27
所以 x^2*(1-x^2)^2<=4/27
x*(1-x^2)<=2/√27
1/(x(1-x^2))>=3√3/2
x/(1-x^2)>=3/2*√3*x^2
同理有
y/(1-y^2)>=3/2*√3*y^2
z/(1-z^2)>=3/2*√3*z^2
三式相加就有：
x/(1-x^2)+y/(1-y^2)+z/(1-z^2)>=3/2*√3*(x^2+y^2+z^2)=3/2*√3
从而得证.