tugiathietvabdtA=1a2+1b2+1c2≥1ab+1bc+1actaco28A−2013≤4A<=>A≤201324
apdungbdtBCScoP≤3(15a2+2ab+b2+15b2+2bc+c2+15c2+2ac+a2)
15a2+2ab+b2≤14(14a2+1(a+b)2)≤116a2+116ab
tudotuongtuvs2hangtukiadk
P≤116(1a2+1b2+1c2+1ab+1bc+1ac)=A16+28A−20134.16=A2+201364≤469764
vayPmax=469764datdkkhia=b=c=√20136√2