Question

已知ab∈R+,并且a≠b,求证a3/b2+b3/a2＞a+b

Answer 1

a3/b2+b3/a2-(a+b)

=a3/b2-b+b3/a2-a

=(a3-b3)/b2+(b3-a3)/a2

=(a3-b3)/b2-(a3-b3)/a2

=(a3-b3)(1/b2-1/a2)

=(a3-b3)(a2-b2)/(a2*b2)

=(a-b)(a2+ab+b2)(a-b)(a+b)/(a2*b2)

=(a-b)2*(a2+ab+b2)(a+b)/(a2*b2)

因为ab∈R+,并且a≠b,

所以a>0,b>0,(a-b)2>0,

从而

a3/b2+b3/a2-(a+b)

=(a-b)2*(a2+ab+b2)(a+b)/(a2*b2)

>0

所以

a3/b2+b3/a2＞a+