$\begin{array}{l} \dfrac{1}{{1 + a}} \ge \left( {1 - \dfrac{1}{{1 + b}}} \right) + \left( {1 - \dfrac{1}{{1 + c}}} \right) = \dfrac{b}{{1 + b}} + \dfrac{c}{{1 + c}} \ge 2\sqrt {\dfrac{{bc}}{{\left( {1 + b} \right)\left( {1 + c} \right)}}} \\ TT:\dfrac{1}{{1 + b}} \ge 2\sqrt {\dfrac{{ca}}{{\left( {1 + a} \right)\left( {1 + c} \right)}}} ,\dfrac{1}{{1 + c}} \ge 2\sqrt {\dfrac{{ba}}{{\left( {1 + b} \right)\left( {1 + a} \right)}}} \\ \Rightarrow \left( {\dfrac{1}{{1 + a}}} \right)\left( {\dfrac{1}{{1 + b}}} \right)\left( {\dfrac{1}{{1 + c}}} \right) \ge 8\sqrt {\dfrac{{{{\left( {abc} \right)}^2}}}{{{{\left[ {\left( {1 + a} \right)\left( {1 + b} \right)\left( {1 + c} \right)} \right]}^2}}}} \\ \Rightarrow 1 \ge 8abc \to abc \le \dfrac{1}{8}\\ \Rightarrow a = b = c = \dfrac{1}{2} \end{array}$
