設(shè)g（x）=f(x)-2/3 x³=1/2 x²+lnx-2/3 x³
即證明當(dāng)x＞1時(shí),g（x）＞0
g(x)′=x+1/x -2x²
=(x²+1-2x³)/x
=(x²-2x+1+2x-2x³)/x
=[(x-1)²-2x(x+1)(x-1)]/x
=-(2x²+x+1)(x-1)/x
顯然當(dāng)x≥1時(shí),g(x)′＞0,g(x)遞增
所以當(dāng)x＞1時(shí),g(x)＞g(1)=1+1-2=0