Bất đẳng thức \({a^2} + {b^2} + {c^2} + {d^2} + {e^2} \ge a(b + c + d + c)\forall \) a, b, c, d, e.

A. \({\left( {a - \frac{b}{2}} \right)^2} + {\left( {a - \frac{c}{2}} \right)^2} + {\left( {a - \frac{d}{2}} \right)^2} + {\left( {a - \frac{e}{2}} \right)^2} \ge 0\)

B. \({\left( {b - \frac{a}{2}} \right)^2} + {\left( {c - \frac{a}{2}} \right)^2} + {\left( {d - \frac{a}{2}} \right)^2} + {\left( {e - \frac{a}{2}} \right)^2} \ge 0\)

C. \({\left( {b + \frac{a}{2}} \right)^2} + {\left( {c + \frac{a}{2}} \right)^2} + {\left( {d + \frac{a}{2}} \right)^2} + {\left( {e + \frac{a}{2}} \right)^2} \ge 0\)

D. \({\left( {a - b} \right)^2} + {\left( {a - c} \right)^2} + {\left( {a - d} \right)^2} + {\left( {a - e} \right)^2} \ge 0\)