Đáp án:

$\begin{array}{l}
1)3\left( {{a^2} + {b^2} + {c^2}} \right) \ge {\left( {a + b + c} \right)^2}\\
 \Rightarrow 3{a^2} + 3{b^2} + 3{c^2} \ge {a^2} + {b^2} + {c^2} + 2ab + 2ac + 2bc\\
 \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca \ge 0\\
 \Rightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right)\\
 + \left( {{c^2} - 2ac + {a^2}} \right) \ge 0\\
 \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0\left( {tm} \right)\\
2){\left( {a + b + c} \right)^2} \ge 3\left( {ab + bc + ca} \right)\\
 \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\\
 \ge 3ab + 3bc + 3ca\\
 \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca \ge 0\\
 \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca \ge 0\\
 \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0\left( {tm} \right)\\
3){a^2} + {b^2} + {c^2} \ge ab + bc + ca\\
 \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca \ge 0\\
 \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0\left( {tm} \right)\\
4)\\
{a^2} + {b^2} + 1 \ge ab + a + b\\
 \Rightarrow 2{a^2} + 2{b^2} + 2 - 2ab - 2a - 2b \ge 0\\
 \Rightarrow \left( {{a^2} - 2a + 1} \right) + \left( {{a^2} - 2ab + {b^2}} \right)\\
 + \left( {{b^2} - 2b + 1} \right) \ge 0\\
 \Rightarrow {\left( {a - 1} \right)^2} + {\left( {a - b} \right)^2} + {\left( {b - 1} \right)^2} \ge 0\left( {tm} \right)\\
5){a^2} + {b^2} + {c^2} + {d^2} + {e^2} \ge a\left( {b + c + d + e} \right)\\
 \Rightarrow \left( {{b^2} - ab + \dfrac{1}{4}{a^2}} \right) + \left( {{c^2} - ac + \dfrac{1}{4}{a^2}} \right)\\
 + \left( {{d^2} - ad + \dfrac{1}{4}{a^2}} \right) + \left( {{e^2} - ae + \dfrac{1}{4}{a^2}} \right) \ge 0\\
 \Rightarrow {\left( {b - \dfrac{1}{2}a} \right)^2} + {\left( {c - \dfrac{1}{2}a} \right)^2}\\
 + {\left( {d - \dfrac{1}{2}a} \right)^2} + {\left( {e - \dfrac{1}{2}a} \right)^2} \ge 0\left( {tmdk} \right)
\end{array}$