Gỉa sử b là cạnh lớn nhất.
Đặt $\dfrac{c}{b}=x;\dfrac{a}{b}=y$ \Rightarrow $x^2+y^2 \le 1$
Ta cm:
$(a+b)(b+c)(c+a) \ge (4+3\sqrt{2})abc$
\Leftrightarrow $(1+\dfrac{b}{a})(1+\dfrac{c}{b})(1+\dfrac{a}{c}) \ge 4+3\sqrt{2}$
\Leftrightarrow $(1+\dfrac{1}{y})(1+x)(1+\dfrac{y}{x}) \ge 4+3\sqrt{2}$
\Leftrightarrow $2+(x+y)+(\dfrac{1}{x}+\dfrac{1}{y})+(\dfrac{x}{y}+\dfrac{y}{x}) \ge 4+3\sqrt{2}$
\Leftrightarrow $(x+y)+(\dfrac{1}{x}+\dfrac{1}{y})+(\dfrac{x}{y}+\dfrac{y}{x}) \ge 2+3\sqrt{2}$
Có: $(x+y)+(\dfrac{1}{x}+\dfrac{1}{y})+(\dfrac{x}{y}+ \dfrac{y}{x}) \ge 2+(x+y)+ \dfrac{4}{x+y}=2+(x+y)+ \dfrac{4}{2(x+y)}+ \dfrac{4}{2(x+y)} \ge 2+2\sqrt{2}+\sqrt{2}=2+3\sqrt{2}$
\Rightarrow đpcm,
\Rightarrow $P \ge 4+3\sqrt{2}$