已知ab∈R+,并且a≠b,求证a3/b2+b3/a2＞a+b

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+b