Áp dụng bất đẳng thức Cauchy – Schwarz ta có:
\[\sqrt {2017} = \sum\limits_{cyc} {\sqrt {{x^2} + {y^2}} } \le \sqrt {6\left( {{x^2} + {y^2} + {z^2}} \right)} \Rightarrow {x^2} + {y^2} + {z^2} \ge \frac{{2017}}{6}\]
\[y + z \le \sqrt {2\left( {{y^2} + {z^2}} \right)} ;\,\,x + z \le \sqrt {2\left( {{x^2} + {z^2}} \right)} ;\,\,x + y \le \sqrt {2\left( {{x^2} + {y^2}} \right)} \]
Khi đó ta được:
\[P = \sum\limits_{cyc} {\frac{{{x^2}}}{{y + z}}} \ge \sum\limits_{cyc} {\frac{{{x^2}}}{{\sqrt {2\left( {{y^2} + {z^2}} \right)} }}} \ge \frac{{\frac{{2017}}{6} - \left( {{y^2} + {z^2}} \right)}}{{\sqrt {2\left( {{y^2} + {z^2}} \right)} }} = \sum\limits_{cyc} {\frac{{2017}}{{6\sqrt {2\left( {{y^2} + {z^2}} \right)} }}} - \frac{1}{{\sqrt 2 }}\sum\limits_{cyc} {\sqrt {{y^2} + {z^2}} } \]
\[\mathop \ge \limits^{AM - GM} \frac{{2017.9}}{{6\sum\limits_{cyc} {\sqrt {2\left( {{y^2} + {z^2}} \right)} } }} - \frac{1}{{\sqrt 2 }}\sum\limits_{cyc} {\sqrt {{y^2} + {z^2}} } = \frac{1}{2}\sqrt {\frac{{2017}}{2}} \]