Áp dụng bất đẳng thức Cô-si cho hai số không âm ta có:
(begin{array}{l}a + b + c = a.1 + left( {b + c} right).1 le frac{{{a^2} + 1}}{2} + frac{{{{left( {b + c} right)}^2} + 1}}{2}\ = frac{{{a^2} + 1 + {b^2} + 2bc + {c^2} + 1}}{2} = frac{{2bc + 4}}{2} = bc + 2,,,left( {do,,,{a^2} + {b^2} + {c^2} = 2} right)\ Rightarrow frac{a}{{bc + 2}} le frac{a}{{a + b + c}}.end{array})
Chứng minh tương tự ta có: (left{ begin{array}{l}frac{b}{{ac + 2}} le frac{b}{{a + b + c}}\frac{c}{{ab + 2}} le frac{c}{{a + b + c}}end{array} right..)
( Rightarrow S le frac{{a + b + c}}{{a + b + c}} = 1.)
Dấu “=” xảy ra ( Leftrightarrow left{ begin{array}{l}a = 1\b + c = 1\{a^2} + {b^2} + {c^2} = 2end{array} right. Leftrightarrow left{ begin{array}{l}a = 1\left[ begin{array}{l}left{ begin{array}{l}b = 0\c = 1end{array} right.\left{ begin{array}{l}b = 1\c = 0end{array} right.end{array} right.end{array} right..)
Vậy (Max,S = 1) khi (left( {a;,,b;,,c} right) = left{ {left( {1;,1;,0} right);,,left( {1;,0;,1} right)} right}.)
Chọn C