x∈(0,e] 证明(e^2)(x^2)-(2/5)x>(x+1)lnx 关键是x不能带零很麻烦啊

设f(x)=(e^2)(x^2)-(2/5)x-(x+1)lnx ,x∈（0,e],
则f'(x)=2e^2*x-7/5-lnx-1/x,
f''(x)=(2e^2*x^2-x+1)/x^2>0,
∴f'(x)↑,
f'(0.27)=0.1957,
f'(0.26)=-0.05677,
∴f(x)|min≈f(0.26)≈2.09 >0,
∴命题成立