ta có 3(a2+b2+c2)=(a+b+c)(a2+b2+c2)=a3+b3+c3+a2b+b2c+c2a+ab2+bc2+ca2 a3+ab2≥2a2b,b3+bc2≥2b2c,c3+ca2≥2c2a
⇒3(a2+b2+c2)≥3(a2b+b2c+c2a)⇒a2+b2+c2≥a2b+b2c+c2a
mà 3(a2+b2+c2)≥(a+b+c)2=9(theobunhiacopxki)⇒a2+b2+c2≥3.
TT cm a4+b4+c4≥(a2+b2+c2)23
⇒P≥7(a2+b2+c2)23+ab+bc+caa2+b2+c2
=7(a2+b2+c2)23+9−(a2+b2+c2)2(a2+b2+c2)
đặt t=a2+b2+c2,t≥3
khi đó P≥7t23+9t2−12 =t212+94t+94t+9t24−12
≥33√t21294t94t+9.94−12=22
dấu "="⇔a=b=c=1