a>4,求证：根号(a-4)-根号(a-1)<根号(a-2)-根号(a+1)

∵ [√(a-4) +√(a+1)]^2 -[√(a-1) +√(a-2)]^2
=[2a-3+2√(a-4)(a+1)] - [2a-3+2√(a-1)(a-2)]
=2[√(a-4)(a+1)-√(a-1)(a-2)]
又 [√(a-4)(a+1]^2 - [√(a-1)(a-2)] ^2
=(a-4)(a+1)-(a-1)(a-2)
=-6<0
∴ √(a-4)(a+1) < √(a-1)(a-2)
∴ [√(a-4) +√(a+1)]^2 < [√(a-1) +√(a-2)]^2
∴ √(a-4)-√(a-1) < √(a-2)-(a+1)

根式有理化得
左边=-3/[根号(a-4)+根号(a-1)]
右边=-3/[根号(a-2)+根号(a+1)]
因为[根号(a-4)+根号(a-1)]<[根号(a-2)+根号(a+1)]
所以1/[根号(a-4)+根号(a-1)]>1/[根号(a-2)+根号(a+1)]
所以-3/[根号(a-4)+根号(a-1)]<-3/[根号(a-2)+根号(a+1)]