题目

若a,b,c,x,y,z,>0,x*2+y*2+z*2=1,求f(x,y,z)=a/x+b/y+c/z的最小值.

提问时间：2021-03-29

答案

令F(x,y,z)=0·λ+f(x,y,z)=λ(x*2+y*2+z*2-1)+f(x,y,z)=λ(x*2+y*2+z*2-1)+a/x+b/y+c/z,
则
偏F/偏λ=x*2+y*2+z*2-1=0; ①
偏F/偏x=2λ·x-a/x^2; ②
偏F/偏y=2λ·y-b/y^2; ③
偏F/偏z=2λ·z-c/z^2; ④
令②③④=0,则
x=[a/(2λ)]^(1/3)
y=[b/(2λ)]^(1/3)
z=[c/(2λ)]^(1/3);
→与①一起解得λ=(1/2)·[a^(2/3) +b^(2/3) +c^(2/3) ]
则代入得
x=a^(1/3)·[a^(2/3) +b^(2/3) +c^(2/3) ]^(-1/2)
y=b^(1/3)·[a^(2/3) +b^(2/3) +c^(2/3) ]^(-1/2)
z=c^(1/3)·[a^(2/3) +b^(2/3) +c^(2/3) ]^(-1/2)
即当x,y,z分别取上述值时f(x,y,z)取得最小值,为
[a^(2/3) +b^(2/3) +c^(2/3) ]^(3/2)