- 均值不等式比较大小问题
- 解 P为最小。在a>b>c>1条件下,首先确定:
(1),M>N <==> a-√c>a-√b <==>√b>√c.
其次确定:
(2)Q>p,<==> a+c+c-3(abc)^(1/3)>a+b-2(ab)^1/2)
<==> c+2(ab)^(1/2)>3(abc)^(1/3)
<==>c+(ab)^(1/2)+(ab)^(1/2)>3(abc)^(1/3) [三元均值不等式]
显然成立.
最后确定:
(3),N>P,<==>a-√b>a+b-2√(ab)<==> 2√(ab)>b+√b.
所以说P为最小。