Question

不等式证明题 证明当x>0时,(x^2-1)Inx>=(x-1)^2

Answer 1

简单计算一下即可，答案如图所示

Answer 2

当x=1时,左边=0=右边.
当0<x =(x-1)^2则,(x+1)Inx<=(x-1)令f(x)=(x+1)Inx-(x-1)则f'(x)=Inx+1/x在令g(x)=Inx+1/x则g'(x)=1/x-1/x^2=(x-1)/x^2<0所以g(x)在0<x g(1)=1,f'(x)>1>0所以f(x)在0<x<1内是增函数即f(x)<f(1)=0即(x+1)inx-(x-1)<0,所以(x+1)inx =(x-1)^2
当x>1时要证(x^2-1)Inx>=(x-1)^2即证(x+1)Inx>=(x-1)同理：令F(x)=(x+1)Inx-(x-1)则由上可得F'(x)=Inx+1/x>0所以F(x)在x>1内是增函数即F(x)>F(1)=0,(x+1)Inx>(x-1)成立.即(x^2-1)Inx>=(x-1)^2
综合可得当x>0时,(x^2-1)Inx>=(x-1)^2</x<1内是增函数即f(x)<f(1)=0即(x+1)inx-(x-1)<0,所以(x+1)inx </x </x