展开方程化简得3x²-2(a+b+c)x+ac+bc+ab=0
判别式△=4(a+b+c)²-4*3(ac+bc+ab)
=4(a²+b²+c²+2ab+2ac+2bc)-12(ac+bc+ab)
=4(a²+b²+c²-ab-ac-bc)
=2(2a²+2b²+2c²-2ab-2ac-2bc)
=2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=2[(a-b)²+(a-c)²+(b-c)²]≥0
所以对于任意实数a,b,c,方程有实根
展开方程化简得3x²-2(a+b+c)x+ac+bc+ab=0
判别式△=4(a+b+c)²-4*3(ac+bc+ab)
=4(a²+b²+c²+2ab+2ac+2bc)-12(ac+bc+ab)
=4(a²+b²+c²-ab-ac-bc)
=2(2a²+2b²+2c²-2ab-2ac-2bc)
=2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=2[(a-b)²+(a-c)²+(b-c)²]≥0
所以对于任意实数a,b,c,方程有实根