解.原不等式可转变为a²+b²+c²+43-ab-9b-8c≤0(a²-ab+0.25b²)+(c²-8c+16)+0.75(b²-12b+36)≤0(a-0.5b)²+(c-4)²+0.75(b-6)²≤0则b=6,c=4,a=0.5b=3注：不等式应该是...