[Toán 9]

L

lp_qt

$-1 \le a;b;c \le 3$

\Leftrightarrow $\left\{\begin{matrix}(a+1)(b+1)(c+1) \ge 0 & \\ (a-3)(b-3)(c-3) \le 0 &
\end{matrix}\right.$

\Leftrightarrow $\left\{\begin{matrix}abc+(ab+bc+ac)+(a+b+c)+1 \ge 0 & \\ abc +9(a+b+c)-3(ab+bc+ac)-27 \le 0 & \end{matrix}\right.$

\Leftrightarrow $\left\{\begin{matrix}abc+(ab+bc+ac)+ 3 +1 \ge 0 & \\ -abc + 3(ab+bc+ac)+27 -9.3 \ge 0 & \end{matrix}\right.$

\Rightarrow $4(ab+bc+ac) +4 \ge 0$

\Leftrightarrow $ab+bc+ac \ge -1$

$(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ac)$

\Rightarrow $a^2+b^2+c^2=9-2(ab+bc+ac) \le 9-2.(-1)=11$

dấu= xảy ra khi có 1 số bằng -1; 1 số bằng 3; 1 số bằng 1.
 
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