Đáp án:
Giải thích các bước giải:
$P=\dfrac{b(a+c)+ac(1-b)}{a+2b+c}=\dfrac{b(a+c)+ac(a+c)}{a+2b+c}=\dfrac{(b+ac)(a+c)}{a+2b+c}$
$⇔P=\dfrac{[b(a+b+c)+ac](a+c)}{a+2b+c}=\dfrac{(a+b)(b+c)(c+a)}{a+2b+c}$
Do $a;b;c \geq 0 ⇒P \geq 0$
$P_{min}=0$ khi $(a;b;c)=(0;0;1)$ và các hoán vị
Lại có: $P=\dfrac{(a+b)(b+c)(c+a)}{a+b+b+c} \leq \dfrac{(a+b)(b+c)(c+a)}{2\sqrt{(a+b)(b+c)}}$
$⇒P \leq \dfrac{1}{2}(a+c)\sqrt{(a+b)(b+c)}=\sqrt{\left(\dfrac{a+c}{2}\right)\left(\dfrac{a+c}{2}\right)(a+b)(b+c)}$
$⇒P \leq \sqrt{\dfrac{1}{256}\left(\dfrac{a+c}{2}+\dfrac{a+c}{2}+a+b+b+c \right)^4}=\sqrt{\dfrac{1}{256}(2a+2b+2c)^4}=\dfrac{1}{4}$
$P_{max}=\dfrac{1}{4}$ khi $(a;b;c)=\left(\dfrac{1}{2};0;\dfrac{1}{2} \right)$
