Đặt [tex]a+b=x,b+c=y,c+a=z\Rightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\Rightarrow \left\{\begin{matrix} \frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\geq 2\sqrt{\frac{yz}{(y+1)(z+1)}}\\ \frac{1}{z+1}=1-\frac{1}{x+1}+1-\frac{1}{y+1}=\frac{x}{x+1}+\frac{y}{y+1}\geq 2\sqrt{\frac{xy}{(x+1)(y+1)}}\\ \frac{1}{y+1}=1-\frac{1}{x+1}+1-\frac{1}{z+1}=\frac{x}{x+1}+\frac{z}{z+1}\geq 2\sqrt{\frac{xz}{(x+1)(z+1)}} \end{matrix}\right.\Rightarrow \frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\geq 8\sqrt{\frac{yz}{(y+1)(z+1)}}.\sqrt{\frac{xy}{(x+1)(y+1)}}.\sqrt{\frac{xz}{(x+1)(z+1)}}=\frac{8xyz}{(x+1)(y+1)(z+1)}\Rightarrow 8xyz\leq 1\Rightarrow xyz\leq \frac{1}{8}\Rightarrow (a+b)(b+c)(c+a)\leq \frac{1}{8}[/tex]
Dấu "=" xảy ra khi [tex]x=y=z\Leftrightarrow a=b=c=\frac{1}{4}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
