设a＞b＞0,求证：√2a^3+3/(ab-b^2)≥10

Question

设a＞b＞0,求证：√2a^3+3/(ab-b^2)≥10

Answer 1

由a>b>0可得0因此原不等式左边
>=√2a^3+3/(a^2/4)
=√2a^3+12/(a^2)
=√2/2a^3+√2/2a^3+4/(a^2)+4/(a^2)+4/(a^2)
>=5*(√2/2a^3*√2/2a^3*(4/(a^2))*(4/(a^2)*(4/(a^2))^(1/5)
=5*(1/2*4*4*4)^(1/5)
=5*2
=10
等号成立当且仅当√2/2a^3=4/a^2,a=(4√2)^(1/5).(2)
由(1),(2),原不等式成立,等号成立当且仅当a=(4√2)^(1/5),b = a/2