Chứng minh $xy^{3}$ + $yz^{3}$ + $zx^{3}$ $\geq$ $xz^{3}$ + $zy^{3}$ + $yx^{3}$ với x $\geq$y$\geq$z $\geq$ 0

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1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   \left( {x{y^3} + y{z^3} + z{x^3}} \right) – \left( {x{z^3} + z{y^3} + y{x^3}} \right)\\
    = \left( {x{y^3} – x{z^3}} \right) + \left( {y{z^3} – z{y^3}} \right) + \left( {z{x^3} – y{x^3}} \right)\\
    = x\left( {{y^3} – {z^3}} \right) + yz\left( {{z^2} – {y^2}} \right) + {x^3}\left( {z – y} \right)\\
    = x\left( {y – z} \right)\left( {{y^2} + yz + {z^2}} \right) + yz\left( {z – y} \right)\left( {z + y} \right) + {x^3}\left( {z – y} \right)\\
    = \left( {y – z} \right)\left( {x{y^2} + xyz + x{z^2} – y{z^2} – {y^2}z – {x^3}} \right)\\
    = \left( {y – z} \right)\left[ {\left( {x{y^2} – {x^3}} \right) + \left( {xyz – {y^2}z} \right) + \left( {x{z^2} – y{z^2}} \right)} \right]\\
    = \left( {y – z} \right)\left[ {x\left( {{y^2} – {x^2}} \right) + yz\left( {x – y} \right) + {z^2}\left( {x – y} \right)} \right]\\
    = \left( {y – z} \right)\left[ {\left( {x – y} \right)\left( {yz + {z^2} – xy – {x^2}} \right)} \right]\\
    = \left( {y – z} \right)\left( {x – y} \right)\left[ {\left( {yz – xy} \right) + \left( {{z^2} – {x^2}} \right)} \right]\\
    = \left( {y – z} \right)\left( {x – y} \right)\left[ {y\left( {z – x} \right) + \left( {z – x} \right)\left( {z + x} \right)} \right]\\
    = \left( {y – z} \right)\left( {x – y} \right)\left( {z – x} \right)\left( {x + y + z} \right) \le 0,\,\,\,\,\forall x \ge y \ge z \ge 0
   \end{array}\)

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