f′（x）=（x-b)²+2(x-a)(x-b)=(x-b)[(x-b)+2(x-a)]
f(x)≥mxf′(x)
→(x-a)(x-b)²≥mx(x-b)[(x-b)+2(x-a)]
→(x-b)﹛(x-a)(x-b)-mx[(x-b)+2(x-a)]﹜≥0
→(x-b)[x²-(a+b)x+ab-mx²+mbx-2mx²+2amx]≥0
→(x-b)[(1-3m)x²+(2am+bm-a-b)x+ab]≥0
→(1-3m)x³+(2am+bm-a-b)x²+abx-(1-3m)bx²-(2am+bm-a-b)bx-ab²≥0
→(1-3m)x³+[2am-a-2b+4bm]x²+(2ab-2abm-b²m+b²)x-ab²≥0
令g(x)=(1-3m)x³+(2am-a-2b+4bm)x²+(2ab-2abm-b²m+b²)x-ab²
m≠1/3時(shí)
g′(x)=(3-9m)x²+(4am-2a-4b+8bm)x+(2ab-2abm-b²m+b²)
欲使g(x)≥0恒成立
→g(x)min≥0
→觀察g′(x):
如果g′(x)恒≥0,即g(x)單調(diào)遞增,此時(shí)g(x)值域肯定為R
g′(x)恒≤0,即g(x)單調(diào)遞減,此時(shí)g(x)值域肯定也為R
如果g′(x)與x軸存在兩個(gè)交點(diǎn),無(wú)論開(kāi)口向上還是向下,g(x)肯定是先增后減再增,或者先減后增再減,這時(shí)g(x)依然沒(méi)有最小值
∴m=1/3