分类：教育/科学-学习帮助
问题：数学:不等式已知a&gt;b&gt;0.求证：(a
已知a>b>0.求证：(a-b)^2/8a<(a+b)/2-√ab<(a-b)^2/8b
回答：已知a>b>0.求证：(a-b)^2/8a<(a+b)/2-√ab<(a-b)^2/8b
证:
∵ b<a ==>a+b<2a ==>a+2(ab)^0.5+b<4a ==>(a^0.5+b^0.5)^2/4a<1 
===>(a^0.5-b^0.5)^2*(a^0.5+b^0.5)^2/(2*4a)<(a^0.5-b^0.5)^2/2
∴ (a-b)^2/8a<(a+b)/2-(ab)^0.5
∵ b<a ==>2b<a+b ==>4b<a+2(ab)^0.5+b ==>1<(a^0.5+b^0.5)^2/4b
===>(a^0.5-b^0.5)^2/2<(a^0.5-b^0.5)^2*(a^0.5+b^0.5)^2/(2*4b)
∴ (a+b)/2-(ab)^0.5<(a-b)^2/8b
∴ (a-b)^2/8a<(a+b)/2-(ab)^0.5<(a-b)^2/8b
证毕: