$\begin{array}{l}Ta\,\,có:\\ a +b + c \leq \dfrac32\\ \to 3\sqrt[3]{abc} \leq a +b + c \leq \dfrac{3}{2}\\ \to \sqrt[3]{abc} \leq \dfrac12\\ \to abc \leq \dfrac18\\ \text{Áp dụng bất đẳng thức AM-GM ta được:}\\ 3 + \dfrac1a + \dfrac1b = 1 + 1 + 1 + \dfrac{1}{2a} + \dfrac{1}{2a} + \dfrac{1}{2b} + \dfrac{1}{2b}\\ \geq 7\sqrt[7]{1\cdot1\cdot1\cdot\dfrac{1}{2a}\cdot\dfrac{1}{2a}\cdot\dfrac{1}{2b}\cdot\dfrac{1}{2}}=7\sqrt[7]{\dfrac{1}{16a^2b^2}}\\ \text{Tương tự ta được:}\\ 3 + \dfrac1b + \dfrac1c \geq 7\sqrt[7]{\dfrac{1}{16b^2c^2}}\\ 3 + \dfrac1c + \dfrac1a \geq 7\sqrt[7]{\dfrac{1}{16c^2a^2}}\\ \text{Nhân vế theo vế ta được:}\\ \left(3 + \dfrac1a + \dfrac1b\right)\cdot\left(3 + \dfrac1b + \dfrac1c\right)\cdot\left(3 + \dfrac1c + \dfrac1a\right) \geq 343\sqrt[7]{\dfrac{1}{16^3.(abc)^4}}\\ \to P \geq 343\sqrt[7]{\dfrac{1}{16^3\cdot\dfrac{1}{8^4}}} = 343\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow a = b = c= \dfrac12\\ \end{array}$
