Cho các số thực x, y, z thỏa mãn căn (log2 x/4)+căn (log3 y/9)+ căn( log5 z/25)=3 . Tính giá trị nhỏ nhất của S= log2011x.log2018 y. log2019 z

Đáp án A

Điều kiện: $x\ge 4;y\ge 9;z\ge 25.$

Đặt $\left\{\begin{array}{l}a=\sqrt{{\log }_2\frac{x}{4}}\Rightarrow a^2={\log }_2\frac{x}{4}\Rightarrow a^2={\log }_{2}x-2\Rightarrow {\log }_{2}x=a^2+2\\ b=\sqrt{{\log }_3\frac{y}{9}}\Rightarrow {\log }_{3}y=b^2+2\\ c=\sqrt{{\log }_5\frac{z}{25}}\Rightarrow {\log }_{5}z=c^2+2\end{array}\right.$

Khi đó $a,b,b\ge 0$  và $a+b+c=3$

Ta có: $\left\{\begin{array}{l}{\log }_{2001}x={\log }_{2001}2.{\log }_{2}x=\left(a^2+2\right).{\log }_{2001}2\\ {\log }_{2018}y=\left(b^2+2\right).{\log }_{2018}3\\ {\log }_{2019}z=\left(c^2+2\right).{\log }_{2019}5\end{array}\right.$

Suy ra $S=\underset{P}{\underbrace{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+1\right)}}.{\log }_{2001}2.{\log }_{2018}3.{\log }_{2019}5.$

Ta có: $\left(a^2+2\right)\left(b^2+2\right)=\left(a^2+1\right)\left(1+b^2\right)+a^2+b^2+3\ge {\left(a+b\right)}^2+\frac{{\left(a+b\right)}^2}{2}+3$

(Bunhiacopxki)

$\begin{array}{l}\Rightarrow \left(a^2+2\right)\left(b^2+2\right)\ge \frac{3}{2}{\left(a+b\right)}^2+3=3\left[\frac{1}{2}{\left(a+b\right)}^2+1\right]\\ \Rightarrow P=\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\ge 3\left[\frac{1}{2}{\left(a+b\right)}^2+1\right]\left(c^2+2\right)\\ =3\left[1+\frac{{\left(a+b\right)}^2}{4}+\frac{{\left(a+b\right)}^2}{4}\right]\left(c^2+1+1\right)\ge 3{\left(c+\frac{a+b}{2}+\frac{a+b}{2}\right)}^2=3{\left(a+b+c\right)}^2=27\end{array}$

 $P=27$ khi a=b=c hay x=8, y=27, z=125

Suy ra $S_{\min }=27.{\log }_{2001}2.{\log }_{2018}3.{\log }_{2019}5$