题目内容
已知a>0,b>0,m>0,n>0.求证:am+n+bm+n≥ambn+anbm.
证明:am+n+bm+n-(ambn+anbm)
=(am+n-ambn)-(anbm-bm+n)
=am(an-bn)-bm(an-bn)
=(am-bm)(an-bn).
当a>b时,am>bm,an>bn,所以(am-bm)(an-bn)>0;
当a<b时,am<bm,an<bn,所以(am-bm)(an-bn)>0;
当a=b时,am=bm,an=bn,所以(am-bm)(an-bn)=0.
综上可知,(am-bm)(an-bn)≥0.
所以am+n+bm+n≥ambn+anbm.
点评:对m、n取具体特殊值,可得到以下一些大家比较熟悉的题目.
(1)已知a>0,b>0.求证:a5+b5>a3b2+a2b3.
(2)已知a>0,b>0.求证:a3+b3≥a2b+b
(3)已知a>0,b>0.求证:a4+b4≥a3b+b
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