Cho {a^2} + {b^2} + {c^2} = ab + bc + ca\) và a + b + c = 2022

Ta có \({a^2} + {b^2} + {c^2} = ab + bc + ca\)

\(2{a^2} + 2{b^2} + 2{c^2} = 2ab + 2bc + 2ca\)

\(2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0\)

\({\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {a - c} \right)^2} = 0\)

Ta có: \({\left( {a - b} \right)^2} \ge 0;\,{\left( {b - c} \right)^2} \ge 0;{\left( {a - c} \right)^2} \ge 0\) nên \({\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {a - c} \right)^2} \ge 0\)

Để \({\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {a - c} \right)^2} = 0\) thì

\[\left\{ \begin{array}{l}{\left( {a - b} \right)^2} = 0\\{\left( {b - c} \right)^2} = 0\\{\left( {a - c} \right)^2} = 0\end{array} \right.\] nên \[\left\{ \begin{array}{l}a - b = 0\\b - c = 0\\a - c = 0\end{array} \right.\] suy ra \[\left\{ \begin{array}{l}a = b\\b = c\\a = c\end{array} \right.\] hay \[a = b = c\].

Do đó \(a + b + c = 2022\).

Vậy \[a = b = c = \frac{{2\,\,022}}{3} = 674\].