ta có 14(a+b+c)3≤(a+b)3+c3≤4(a3+b3)+c3≤2(a+b+c)((a+b+c)24−2⇒a+b+c≥42a23a2+b2+2a(c+2)=aa+c+2+b22a+a2≤aa+c+2+2√b22aa2=aa+b+c+2có (a+b)2+c2≥12(a+b+c)2
khi đó P ≤a+b+ca+b+c+2−(a+b+c)232 Đặt t=a+b+c≥4
⇒P≤tt+2−t232=f(t)
f′(t)=2(t+2)2−t16<0,∀t≥4
\Rightarrow f(t) nb/ \left[ 4{;} +vô cùng\right]
\Rightarrow P \leq f(t) \leq f(4)=\frac{1}{6}
dấu "=" \Leftrightarrow a=b=1;c=2