已知x1,x2为R+,4^X=(1+f(x)\=(1-f(x))且f(x1)+f(x2)=1求f(X1+x2)的min

已知x1,x2为R+,4^X=(1+f(x)\=(1-f(x))且f(x1)+f(x2)=1求f(X1+x2)的min
已知x1,x2为R+,4^X=(1+f(x)\=(1-f(x))且f(x1)+f(x2)=1求f(X1+x2)的min

已知x1,x2为R+,4^X=(1+f(x)\=(1-f(x))且f(x1)+f(x2)=1求f(X1+x2)的min
^x=[1+f(x)]/[1-f(x)]
---->f(x)=[1-4^x]/[1+4^x]
设a=4^(x1),b=4^(x2),显然a>0,b>0.
f(x1)+f(x2)=(1-a)/(1+a)+(1-b)/(1+b)=(2-2ab)/[(1+a)(1+b)]=1
---->2-2ab=(1+a)(1+b)=1+a+b+ab
--->3ab+a+b-1=0
因为a+b>=2√(ab),所以
3ab+2√(ab)-1