Có (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)≥a3+b3+c3+24abc
⇒P≥ab+bc+caa2+b2+c2+a3+b3+c3abc+24
≥33√a2b2c2a2+b2+c2+a3+b3+c33abc+a3+b3+c33abc+a3+b3+c33abc+24
≥44√(a3+b3+c3)39(a2+b2+c2).3√a7b7c7+24
≥44√(a3+b3+c3)2(a+b+c)273√a7b7c7+24 ( Do a3+b3+c3a2+b2+c2≥a+b+c3)
≥44√9a2b2c2.33√abc273√a7b7c7+24
=28