\(P=\frac{x^{3} z}{y^{2} \left(xz+y^{2} \right)} +\frac{y^{4} }{z^{2} \left(xz+y^{2} \right)} +\frac{z^{3} +15x^{3} }{x^{2} z} \)
\(P=\frac{x^{3} z}{y^{2} xz+y^{4} } +\frac{y^{4} }{z^{3} x+z^{2} y^{2} } +\frac{z^{2} }{x^{2} } +\frac{15x}{z} \)
\(P=\frac{\left(\frac{x}{y} \right)^{3} }{\frac{x}{y} +\frac{y}{z} } +\frac{\left(\frac{y}{z} \right)^{3} }{\frac{x}{y} +\frac{y}{z} } +\left(\frac{z}{x} \right)^{2} +\frac{15}{\left(\frac{z}{x} \right)} \)
Đặt \(a=\frac{x}{y} ,b=\frac{y}{z} , c=\frac{z}{x} với 0<a<1;\, 0<b<1;\, c>1 \Rightarrow a.b.c=1\Rightarrow ab=\frac{1}{c}\)
\(P=\frac{a^{3} }{a+b} +\frac{b^{3} }{a+b} +c^{2} +\frac{15}{c} =\frac{a^{3} +b^{3} }{a+b} +c^{2} +\frac{15}{c} \)
Ta có \(\left(a+b\right)\left(a-b\right)^{2} \ge 0\Leftrightarrow \left(a^{2} -b^{2} \right)\left(a-b\right)\ge 0\Leftrightarrow a^{2} \left(a-b\right)-b^{2} \left(a-b\right)\ge 0 \)
\(\Leftrightarrow a^{3} +b^{3} -a^{2} b-ab^{2} \ge 0\Leftrightarrow a^{3} +b^{3} \ge ab\left(a+b\right)\)
\(\Rightarrow \frac{a^{3} +b^{3} }{a+b} \ge ab=\frac{1}{c} \Rightarrow P\ge ab+c^{2} +\frac{15}{c} \Rightarrow P\ge \frac{1}{c} +c^{2} +\frac{15}{c} =\frac{16}{c} +c^{2} \)
Khảo sát hàm số \(P=f\left(c\right)=\frac{16}{c} +c^{2} \)
\(P'=f'\left(c\right)=\frac{-16}{c^{2} } +2c=\frac{-16+2c^{3} }{c^{2} } \)
\(P'=0\Leftrightarrow 2c^{3} -16=0\Leftrightarrow c^{3} =8\Leftrightarrow c=2>1\) (thỏa mãn)
Bảng biến thiên
\(\Rightarrow {\mathop{\min }\limits_{{\rm R}}} P=12 khi c=2.ab=\frac{1}{c} =\frac{1}{2} \Leftrightarrow \left\{\begin{array}{l} {\frac{z}{x} =2} \\ {\frac{x}{y} .\frac{y}{z} =\frac{1}{2} } \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {z=2x} \\ {\frac{x}{z} =\frac{1}{2} } \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} {z=2x} \\ {z=2x} \end{array}\right. \Leftrightarrow z=2x\)
monmon70023220 
Darling_274 
minhquanhhqt160 
tngnhatganh117 