Cho các số thực x, y, z thỏa mãn $\sqrt{{{\log...

Điều kiện: $x\ge 4;y\ge 9;z\ge 25.$
Đặt $\left\{ \begin{aligned}
& a=\sqrt{{{\log }_{2}}\dfrac{x}{4}}\Rightarrow {{a}^{2}}={{\log }_{2}}\dfrac{x}{4}\Rightarrow {{a}^{2}}={{\log }_{2}}x-2\Rightarrow {{\log }_{2}}x={{a}^{2}}+2 \\
& b=\sqrt{{{\log }_{3}}\dfrac{y}{9}}\Rightarrow {{\log }_{3}}y={{b}^{2}}+2 \\
& c=\sqrt{{{\log }_{5}}\dfrac{z}{25}}\Rightarrow {{\log }_{5}}z={{c}^{2}}+2 \\
\end{aligned} \right.$
Khi đó $a,b,b\ge 0$ và $a+b+c=3$
Ta có: $\left\{ \begin{aligned}
& {{\log }_{2001}}x={{\log }_{2001}}2.{{\log }_{2}}x=\left( {{a}^{2}}+2 \right).{{\log }_{2001}}2 \\
& {{\log }_{2018}}y=\left( {{b}^{2}}+2 \right).{{\log }_{2018}}3 \\
& {{\log }_{2019}}z=\left( {{c}^{2}}+2 \right).{{\log }_{2019}}5 \\
\end{aligned} \right.$
Suy ra $S=\underbrace{\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)\left( {{c}^{2}}+1 \right)}_{P}.{{\log }_{2001}}2.{{\log }_{2018}}3.{{\log }_{2019}}5.$
Ta có: $\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)=\left( {{a}^{2}}+1 \right)\left( 1+{{b}^{2}} \right)+{{a}^{2}}+{{b}^{2}}+3\ge {{\left( a+b \right)}^{2}}+\dfrac{{{\left( a+b \right)}^{2}}}{2}+3$
(Bunhiacopxki)
$\begin{aligned}
& \Rightarrow \left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)\ge \dfrac{3}{2}{{\left( a+b \right)}^{2}}+3=3\left[ \dfrac{1}{2}{{\left( a+b \right)}^{2}}+1 \right] \\
& \Rightarrow P=\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)\left( {{c}^{2}}+2 \right)\ge 3\left[ \dfrac{1}{2}{{\left( a+b \right)}^{2}}+1 \right]\left( {{c}^{2}}+2 \right) \\
& =3\left[ 1+\dfrac{{{\left( a+b \right)}^{2}}}{4}+\dfrac{{{\left( a+b \right)}^{2}}}{4} \right]\left( {{c}^{2}}+1+1 \right)\ge 3{{\left( c+\dfrac{a+b}{2}+\dfrac{a+b}{2} \right)}^{2}}=3{{\left( a+b+c \right)}^{2}}=27 \\
\end{aligned}$
$P=27$ khi $a=b=c=1$ hay $x=8,y=27,z=125$
Suy ra ${{S}_{\min }}=27.{{\log }_{2001}}2.{{\log }_{2018}}3.{{\log }_{2019}}5$