设a、b、c为△ABC三边,证明：a(3a+2b+c)²-2b(b+c) +a-2b-2c≥0.

9a³-2ab²+ac²+12a²b+6a²c-2abc-4b³-6b²c-bc²+a-2b-2c
=9a³+4ab²+ac²+12a²b+6a²c+4abc-6ab²-4b³-2b²c-6abc-4b²c-bc²+a-2b-2c
=a(3a+2b+c)^2-2(b+c)(3a+2b+c)+[a-2(b+c)].①
设3a+2b+c=x,a=A,-2(b+c)=B,[a-2(b+c)]=C
则①为Ax^2+Bx+C,原命题为证明：Ax^2+Bx+C≥0
A>0,△=B^2-4AC=-(b+c)^2+2a(b+c)-a^2
=-[(b+c)^2-2a(b+c)+a^2]
=-[(b+c)-a]^2
∵b+c>a,∴(b+c)-a≠0,∴[(b+c)-a]^2>0,∴-[(b+c)-a]^2