Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {x{y^3} + y{z^3} + z{x^3}} \right) - \left( {x{z^3} + z{y^3} + y{x^3}} \right)\\
= \left( {x{y^3} - x{z^3}} \right) + \left( {y{z^3} - z{y^3}} \right) + \left( {z{x^3} - y{x^3}} \right)\\
= x\left( {{y^3} - {z^3}} \right) + yz\left( {{z^2} - {y^2}} \right) + {x^3}\left( {z - y} \right)\\
= x\left( {y - z} \right)\left( {{y^2} + yz + {z^2}} \right) + yz\left( {z - y} \right)\left( {z + y} \right) + {x^3}\left( {z - y} \right)\\
= \left( {y - z} \right)\left( {x{y^2} + xyz + x{z^2} - y{z^2} - {y^2}z - {x^3}} \right)\\
= \left( {y - z} \right)\left[ {\left( {x{y^2} - {x^3}} \right) + \left( {xyz - {y^2}z} \right) + \left( {x{z^2} - y{z^2}} \right)} \right]\\
= \left( {y - z} \right)\left[ {x\left( {{y^2} - {x^2}} \right) + yz\left( {x - y} \right) + {z^2}\left( {x - y} \right)} \right]\\
= \left( {y - z} \right)\left[ {\left( {x - y} \right)\left( {yz + {z^2} - xy - {x^2}} \right)} \right]\\
= \left( {y - z} \right)\left( {x - y} \right)\left[ {\left( {yz - xy} \right) + \left( {{z^2} - {x^2}} \right)} \right]\\
= \left( {y - z} \right)\left( {x - y} \right)\left[ {y\left( {z - x} \right) + \left( {z - x} \right)\left( {z + x} \right)} \right]\\
= \left( {y - z} \right)\left( {x - y} \right)\left( {z - x} \right)\left( {x + y + z} \right) \le 0,\,\,\,\,\forall x \ge y \ge z \ge 0
\end{array}\)
