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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!
人气:418 ℃ 时间:2020-06-08 00:38:44
解答
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.
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