Cho x+y= 2. CMR : x^2017 + y^2017 bé hơn hoặc bằng x^2018+ y^2018

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1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   \left( {{x^{2018}} + {y^{2018}}} \right) – \left( {{x^{2017}}y + x{y^{2017}}} \right)\\
    = {x^{2018}} + {y^{2018}} – {x^{2017}}y – x{y^{2017}}\\
    = {x^{2017}}\left( {x – y} \right) + {y^{2017}}\left( {y – x} \right)\\
    = \left( {x – y} \right)\left( {{x^{2017}} – {y^{2017}}} \right)\\
    = {\left( {x – y} \right)^2}\left( {{x^{2016}} + {x^{2015}}y + {x^{2014}}{y^2} + …. + x{y^{2015}} + {y^{2016}}} \right) \ge 0,\,\,\,\,\,\forall x,y\\
    \Rightarrow {x^{2018}} + {y^{2018}} \ge {x^{2017}}y + x{y^{2017}}\\
   2\left( {{x^{2017}} + {y^{2017}}} \right) = \left( {x + y} \right)\left( {{x^{2017}} + {y^{2017}}} \right)\\
    = \left( {{x^{2018}} + {y^{2018}}} \right) + \left( {{x^{2017}}y + x{y^{2017}}} \right) \le 2\left( {{x^{2018}} + {y^{2018}}} \right)\\
    \Rightarrow {x^{2017}} + {y^{2017}} \le {x^{2018}} + {y^{2018}}
   \end{array}\)

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