已知abc都是整数,且满足a²+b²+c²+4≤ab+3b+2c,求abc的值 快!

a^2+b^2+c^2+4≤ab+3b+2c
a^2-ab+(1/4)b^2+(3/4)b^2-3b+3+c^2-2c+1≤0
[a^2-ab+(1/4)b^2]+3*[(1/4)b^2-b+1]+[c^2-2c+1]≤0
则 [a-(1/2)b]^2+3*[(1/2)b-1]^2+(c-1)^2