1/a+1/b+k/(a+b)≥0恒成立k/(a+b)≥-(1/a+1/b)k≥-(a+b) *(a+b)/(ab)=-(a+b)²/(ab)恒成立,需 k≥[-(a+b)²/(ab)]的最大值∵a>0,b>0∴a+b≥2√(ab) (a=b时,取等号）两边平方∴(a+b)²≥4ab ∴(a+b)²...