∵a>b>0∴a²＞ab＞0即：a²-ab＞0且ab＞0a² + 1/ab + 1/a(a-b)=a² + 1/ab + 1/(a²-ab) -ab+ab=[(a²-ab)+1/(a²-ab)] + [ab+1/(ab)]≥2+2 【基本不等式】=4當(dāng)且僅當(dāng)a²-ab=1、a...這步{(a²-ab)+[1/(a²-ab)]}+[(ab)+1/(ab)]≥2+2=4。具體怎么得出來的？a² + 1/ab + 1/a(a-b) 加個(gè)ab，減個(gè)ab，等式不變=a² + 1/ab + 1/(a²-ab) -ab+ab 運(yùn)用加法交換律，+ab和1/ab結(jié)合，-ab和a²結(jié)合[(a²-ab)+1/(a²-ab)]+[ab+1/(ab)] 運(yùn)用基本不等式≥2+2“運(yùn)用基本不等式≥2+2”這又是如何得出的？必修5基礎(chǔ)知識，基本不等式：a+b≥2√ab 【√是根號】當(dāng)且僅當(dāng)a=b時(shí)取等號 [(a²-ab)+1/(a²-ab)]+[ab+1/(ab)]≥2√[(a²-ab)×1/(a²-ab)] + 2√[ab×1/(ab)]=2√1 + 2√1=2+2=4