Tìm giá trị nhỏ nhất của $P=\dfrac{{{x}^{3}}z}{{{y}^{2}}\left(xz+{{y}^{2}} \right)}+\dfrac{{{y}^{4}}}{{{z}^{2}}\left( xz+{{y}^{2}}...

Ta có: $P=\dfrac{{{x}^{3}}z}{{{y}^{2}}\left( xz+{{y}^{2}} \right)}+\dfrac{{{y}^{4}}}{{{z}^{2}}\left( xz+{{y}^{2}} \right)}+\dfrac{{{z}^{3}}+15{{x}^{3}}}{{{x}^{2}}z}=\dfrac{{{\left( \dfrac{x}{y} \right)}^{3}}}{\left( \dfrac{x}{y}+\dfrac{y}{z} \right)}+\dfrac{{{\left( \dfrac{y}{z} \right)}^{3}}}{\left( \dfrac{x}{y}+\dfrac{y}{z} \right)}+{{\left( \dfrac{z}{x} \right)}^{2}}+\dfrac{15}{\dfrac{z}{x}}$
Đặt $a=\dfrac{x}{y}<1,b=\dfrac{y}{z}<1,c=\dfrac{z}{x}>1$ và $abc=1\Leftrightarrow ab=\dfrac{1}{c}.$
Ta được: $P=\dfrac{{{a}^{3}}}{\left( a+b \right)}+\dfrac{{{b}^{3}}}{\left( a+b \right)}+{{c}^{2}}+\dfrac{15}{c}={{a}^{2}}+{{b}^{2}}-ab+{{c}^{2}}+\dfrac{15}{c}\ge ab+{{c}^{2}}+\dfrac{15}{c}$
$={{c}^{2}}+\dfrac{16}{c}={{c}^{2}}+\dfrac{8}{c}+\dfrac{8}{c}\ge 3\sqrt[3]{{{c}^{2}}.\dfrac{8}{c}.\dfrac{8}{c}}=12.$
Vậy ${{P}_{\min }}=12$ khi và chỉ khi $\left\{ \begin{aligned}
& a=b \\
& abc=1 \\
& {{c}^{2}}=\dfrac{8}{c} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=b=\dfrac{1}{\sqrt{2}} \\
& c=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=\dfrac{1}{\sqrt{2}}y \\
& y=\dfrac{1}{\sqrt{2}}z \\
& z=2x \\
\end{aligned} \right..$