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

Giải thích các bước giải:

 Ta có:

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