(a²-ab+1/4b²)+(3/4b²-9b+27)+(c²-8c+16)<1
(a²-ab+1/4b²)+3/4(b²-12b+36)+(c²-8c+16)<1
(a-1/2b)²+3/4(b-6)²+(c-4)²<1···········①
由于a、b、c均为正整数,所以
(c-4)²<1,
解得:
-1
此时①式变为:
(a-1/2b)²+3/4(b-6)²<1···············②
可得:
0≤3/4(b-6)²<1
(b-6)²<4/3
解得:
-(2√3)/3
b=6时,代入②式,得:a=3,
b=5、7时,分别代入②式,同时得:
(a-5/2)²+3/4<1
(a-5/2)²<1/4
可得:
-1/2
综上,a、b、c的值为:
a=3
b=6
c=4