已知a>b>0,求证：(a-b)^2/8a < (a+b/2)-跟号ab

题目

已知a>b>0,求证：(a-b)^2/8a < (a+b/2)-跟号ab

答案

a/b >1;
b/a ((√a - √b)^2/2)*4/4
= (a+b)/2-√(ab);
(a-b)^2/(8a) = ((√a - √b)^2/2)*((√a + √b)^2/(4a))
= ((√a - √b)^2/2)*((a+b+2√(ab))/a)/4
= ((a+b)/2-√(ab))*(1+b/a + 2*(√a)/(√b)))/4
< ((a+b)/2-√(ab))*(1+1+2)/4
= (a+b)/2-√(ab)
从而：
(a-b)^2/ 8a