Giả sử \(x = \min \{ x, y, z\}\) suy ra \(x \in \left [ 0;\frac{1}{2} \right ]\)
Ta có \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
\(\Rightarrow x^3 + y^3 + z^3 = 3xyz + (x+y+z)[(x+y+z)^2 - 3(xy + yz + zx)]\)
\(= 3xyz + \frac{27}{8} - \frac{9(xy + yz + zx)}{2}\)
Ta có \(P = x^3 + y^3 + z^3 + x^2y^2z^2 = x^2y^2z^2 + 3xyz + \frac{27}{8} - \frac{9}{2}(xy + yz + zx)\)
\(= \left ( xyz - \frac{1}{8} \right )^2 - \frac{1}{64} + \frac{13}{4}xyz + \frac{27}{8} - \frac{9}{2}(xy + yz + zx)\) \(\geq \frac{215}{64} - \frac{9}{2}(xy + zx) - yz \left ( \frac{9}{2} - \frac{13}{4} \right )\)
Vì \(x \in \left [ 0; \frac{1}{2} \right ] \Rightarrow \frac{9}{2} - \frac{13}{4}x \geq 0 \Rightarrow -yz\left ( \frac{9}{2} - \frac{13}{4}x \right ) \geq \left ( \frac{y+z}{2} \right )^2 \left ( \frac{9}{2} - \frac{13}{4}x \right )\)
Suy ra \(P \geq \frac{215}{64} - \frac{9}{2}x \left (\frac{3}{2} - x \right ) - \frac{1}{4} \left (\frac{3}{2} - x \right )^2 \left (\frac{9}{2} - \frac{13}{4}x \right )\)
Xet\(f(x) = \frac{215}{64} - \frac{9}{2}x \left (\frac{3}{2} - x \right ) - \frac{1}{4} \left (\frac{3}{2} - x \right )^2 \left (\frac{9}{2} - \frac{13}{4}x \right ), x \in \left [ 0; \frac{1}{2} \right ]\)
Hàm số f(x) nghịch biến trên \(\left [ 0; \frac{1}{2} \right ] \Rightarrow f_{(x)} \geq f_{(\frac{1}{2})} = \frac{25}{64}\)
Vậy GTLN của P bằng \(\frac{25}{64}\) đạt khi \(x = y = z = \frac{1}{2}\)
