Question

证明:当x>0时,f(x)=e^x+2x+x²-2(x+1)ln(x+1)>1

Answer 1

显然x=0时，f(0)=1
求导得到f'(x)=e^x+2+2x-2ln(x+1)-2
=e^x+2x-2ln(x+1)
那么f'(0)=1
>0
继续求导f''(x)=e^x+2-2/(x+1)=e^x+2x/(x+1)
显然二阶导数恒大于0
那么f'(x)递增，恒大于零
故f(x)恒大于1

Answer 2

f(x)=e^(x-m)-ln(2x)>=e^(x-2)-ln(2x),即证e^(x-2)-ln(2x)>-ln2,即e^(x-2)>ln(2x)-ln2=1nx,令g(x)=e^(x-2)-1nx,g'(x)=e^(x-2)-1/x,显然g'(x)递增，设g'(x)=e^(x-2)-1/x=0的根为x1,则g(x)在（0,x1)递减，在（x1,+8)递增，故在x=x1取最小值。e^（x1-2)=1/x1,取对数得x1-2=1n(1/x1),即2-x1=1nx1且易知x1>1,故有1nx1=2-x1<1/x1,则e^（x1-2)=1/x1>1nx1，有g(x)>=g(x1)=e^（x1-2)-1nx1>0,故对x>0，都有g(x)>0,即e^(x-2)>1nx=1n2x-1n2,e^（x-2)-1n2x>-1n2,所以当m<=2时，f(x)=e^(x-m)-ln(2x)>=e^(x-2)-ln(2x)>-1n2,即f(x)>-1n2。