若a>0,b>0,a+b=1,则(1/a^2-1)(1/b^2-1)的最小值是?最好写在纸上上传图片谢谢

Question

若a>0,b>0,a+b=1,则(1/a^2-1)(1/b^2-1)的最小值是?
最好写在纸上上传图片
谢谢

Answer 1

(1/a^2-1)(1/b^2-1)
=(1-aa)(1-bb)/(aabb)
=1+(1-aa-bb)/(aabb)
=1+[(a+b)^4-(aa+bb)(aa+2ab+bb)]/aabb
=1+[aaaa+4aaab+6aabb+4abbb+bbbb-aaaa-bbaa-2aaab-2abbb-aabb-bbbb]/aabb
=1+[2aaab+4aabb+2abbb]/aabb
=1+2[aa+2ab+bb]/ab
>=1+8=9