[tex]P^2=2(a+b+c)+3+2(\sqrt{(a+b+1)(b+c+1)}+\sqrt{(b+c+1)(c+a+1)}+\sqrt{(c+a+1)(a+b+1)})=9+2(\sqrt{(a+b+1)(b+c+1)}+\sqrt{(b+c+1)(c+a+1)}+\sqrt{(c+a+1)(a+b+1)})[/tex]
Lại có:[tex]\sqrt{(a+b+1)(b+c+1)}=\sqrt{ab+ac+b^2+bc+b+1+a+b+c}=\sqrt{b(a+b+c)+ac+b+4}=\sqrt{3b+ac+b+4}=\sqrt{ac+4b+4} \geq \sqrt{4b+4}=2\sqrt{b+1}[/tex]
Tương tự ta có: [tex]P^2 \geq 9+4(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})[/tex]
Mà [tex](\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})^2=a+b+c+3+2(\sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)})=6+2(\sqrt{ab+a+b+1}+\sqrt{bc+b+c+1}+\sqrt{ac+a+c+1}) \geq 6+2(\sqrt{a+b+1}+\sqrt{b+c+1}+\sqrt{c+a+1})=6+2P[/tex]
[tex]P^2-9 \geq 4(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}) \Rightarrow P^4-18P^2+81 \geq 16(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})^2 \geq 16(6+2P) \Rightarrow P^4-18P^2-32P-15 \geq 0 \Rightarrow (P-5)(P^3+5P^2+7P+3) \geq 0 \Rightarrow P \geq 5[/tex]
Dấu "=" xảy ra chẳng hạn a = 3, b = c = 0.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
