Giải thích các bước giải:
\(\begin{array}{l}
{a^2} + {b^2} \ge 2ab;\,\,\,\,{b^2} + {c^2} \ge 2bc;\,\,\,\,\,{c^2} + {a^2} \ge 2ca\\
\Rightarrow \left( {{a^2} + {b^2}} \right) + \left( {{b^2} + {c^2}} \right) + \left( {{c^2} + {a^2}} \right) \ge 2ab + 2bc + 2ca \Rightarrow {a^2} + {b^2} + {c^2} \ge ab + bc + ca\\
a + b + c = 3 \Leftrightarrow {\left( {a + b + c} \right)^2} = 9\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 9\\
\Rightarrow 9 \le 3\left( {{a^2} + {b^2} + {c^2}} \right) \Rightarrow {a^2} + {b^2} + {c^2} \ge 3
\end{array}\)
Áp dụng BĐT Bunhia- copski \(\left( {\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{x^2} + {y^2} + {z^2}} \right) \ge {{\left( {ax + by + cz} \right)}^2}} \right)\), ta có:
\(\begin{array}{l}
a,\\
3\left( {{a^3} + {b^3} + {c^3}} \right)\\
= \left( {a + b + c} \right)\left( {{a^3} + {b^3} + {c^3}} \right)\\
\ge {\left( {\sqrt a .\sqrt {{a^3}} + \sqrt b .\sqrt {{b^3}} + \sqrt c .\sqrt {{c^3}} } \right)^2}\\
= {\left( {{a^2} + {b^2} + {c^2}} \right)^2}\\
3 \le {a^2} + {b^2} + {c^2} \Rightarrow {a^3} + {b^3} + {c^3} \ge {a^2} + {b^2} + {c^2}\\
b,\\
\left( {{a^4} + {b^4} + {c^4}} \right)\left( {{a^2} + {b^2} + {c^2}} \right)\\
\ge {\left( {{a^2}.a + {b^2}.b + {c^2}.c} \right)^2} = {\left( {{a^3} + {b^2} + {c^3}} \right)^2}\\
{a^2} + {b^2} + {c^2} \le {a^3} + {b^3} + {c^3} \Rightarrow {a^4} + {b^4} + {c^4} \ge {a^3} + {b^3} + {c^3}
\end{array}\)
