两式相减(a/b2+b/a2)-(1/a+1/b)
=(a3+b3)/(ab)2-(a+b)/ab
通分 =(a3+b3)/(ab)2-(a2b+b2a)/(ab)2
=[(a3+b3)-(a2b+b2a)]/(ab)2
=[a2(a-b)-b2(a-b)]/(ab)2
=[(a+b)(a-b)2]/(ab)2
因为a+b>0,(a-b)2≥0.(ab)2>0
所以 (a/b2+b/a2)-(1/a+1/b)≥0
即当a=b时两式相等
a≠b时 (a/b2+b/a2)>(1/a+1/b)
两式相减(a/b2+b/a2)-(1/a+1/b)
=(a3+b3)/(ab)2-(a+b)/ab
通分 =(a3+b3)/(ab)2-(a2b+b2a)/(ab)2
=[(a3+b3)-(a2b+b2a)]/(ab)2
=[a2(a-b)-b2(a-b)]/(ab)2
=[(a+b)(a-b)2]/(ab)2
因为a+b>0,(a-b)2≥0.(ab)2>0
所以 (a/b2+b/a2)-(1/a+1/b)≥0
即当a=b时两式相等
a≠b时 (a/b2+b/a2)>(1/a+1/b)