Lời giải:

Ta có: xy + yz + xz = 1 \( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{{x^2} + 1 = {x^2} + xy + yz + xz = \left( {x + y} \right)\left( {x + z} \right)}\\{{y^2} + 1 = {y^2} + xy + yz + xz = \left( {y + z} \right)\left( {y + x} \right)}\\{{z^2} + 1 = {z^2} + xy + yz + xz = \left( {z + x} \right)\left( {z + y} \right)}\end{array}} \right.\)

Do đó: \(\sqrt {\frac{{\left( {{y^2} + 1} \right)\left( {{z^2} + 1} \right)}}{{{x^2} + 1}}} = \sqrt {\frac{{\left( {y + z} \right)\left( {y + x} \right)\left( {z + x} \right)\left( {z + y} \right)}}{{\left( {x + y} \right)\left( {x + z} \right)}}} = \sqrt {{{\left( {y + z} \right)}^2}} = y + z\) (do y, z dương)

\( \Rightarrow x\sqrt {\frac{{\left( {{y^2} + 1} \right)\left( {{z^2} + 1} \right)}}{{{x^2} + 1}}} = x\left( {y + z} \right)\).

Tương tự: \(y\sqrt {\frac{{\left( {{z^2} + 1} \right)\left( {{x^2} + 1} \right)}}{{{y^2} + 1}}} = y\left( {x + z} \right);z\sqrt {\frac{{\left( {{x^2} + 1} \right)\left( {{y^2} + 1} \right)}}{{{z^2} + 1}}} = z\left( {x + y} \right)\).

Do đó: \(A = x\left( {y + z} \right) + y\left( {x + z} \right) + z\left( {x + y} \right) = 2\left( {xy + yz + xz} \right) = 2\).