a,b,c,d都是正数.
a^4 + b^4 + c^4 + d^4 >= 2a^2b^2 + 2c^2d^2 = 2[(ab)^2 + (cd)^2] >= 2[2abcd] = 4abcd
等号成立当且仅当a^2 = b^2并且c^2 = d^2并且ab = cd.
也即,a = b并且c=d并且a^2=c^2.
也即,a = b并且c=d并且a=c.
就是,a = b = c = d.
所以,四边形ABCD是菱形.
a,b,c,d都是正数.
a^4 + b^4 + c^4 + d^4 >= 2a^2b^2 + 2c^2d^2 = 2[(ab)^2 + (cd)^2] >= 2[2abcd] = 4abcd
等号成立当且仅当a^2 = b^2并且c^2 = d^2并且ab = cd.
也即,a = b并且c=d并且a^2=c^2.
也即,a = b并且c=d并且a=c.
就是,a = b = c = d.
所以,四边形ABCD是菱形.