已知a>0,b>0,且a不等于b,比较(a^2/b+b^2/a)与(a+b)的大小

(差法）[(a²/b)+(b²/a)]-(a+b)=[(a²/b)-b]+[(b²/a)-a]=[(a²-b²)/b]+[(b²-a²)/a]=(a²-b²)[(1/b)-(1/a)]=(a+b)(a-b)[(a-b)/(ab)]=(a+b)(a-b)²/(ab).因a,b>0,a≠b.故a+b,ab>0,(a-b)²>0.===>[(a²/b)+(b²/a)]-(a+b)＞0.===>(a²/b)+(b²/a)＞a+b.