Xét các số thực dương $a,b,c$ lớn hơn $1$ ( với $a>b$ ) thỏa mãn $4\left( {{\log }_{a}}c+{{\log }_{b}}c \right)=25{{\log }_{ab}}c$. Giá trị nhỏ...

Đặt \)">{{\log }_{c}}a=x,{{\log }_{c}}b=ya,b,c>1a>b{{\log }_{c}}a>{{\log }_{c}}bx>y>04\left( \dfrac{1}{{{\log }_{c}}a}+\dfrac{1}{{{\log }_{c}}b} \right)=25.\dfrac{1}{{{\log }_{c}}ab}\Leftrightarrow \dfrac{4}{x}+\dfrac{4}{y}=\dfrac{25}{x+y}\Leftrightarrow \dfrac{{{\left( x+y \right)}^{2}}}{xy}=\dfrac{25}{4}\Leftrightarrow \dfrac{x}{y}+\dfrac{y}{x}=\dfrac{17}{4}\Rightarrow \left[ \begin{aligned}
& \dfrac{x}{y}=4 \\
& \dfrac{x}{y}=\dfrac{1}{4} \\
\end{aligned} \right. \Rightarrow x=4y x>y{{\log }_{b}}a+{{\log }_{a}}c+{{\log }_{c}}b=\dfrac{{{\log }_{c}}a}{{{\log }_{c}}b}+\dfrac{1}{{{\log }_{c}}a}+{{\log }_{c}}b=\dfrac{x}{y}+\dfrac{1}{x}+y=\dfrac{x}{y}+\dfrac{1}{4y}+y\ge 4+2\sqrt{\dfrac{1}{4y}.y}=5y=\dfrac{1}{2}x=2a={{c}^{2}};c={{b}^{2}}54\left( {{\log }_{a}}b.{{\log }_{b}}c+{{\log }_{b}}c \right)=25.{{\log }_{ab}}b.{{\log }_{b}}c\Leftrightarrow 4{{\log }_{b}}c\left( {{\log }_{a}}b+1 \right)=25\dfrac{{{\log }_{b}}c}{{{\log }_{b}}ab}\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{b}}c=0 \\
& 4\left( {{\log }_{a}}b+1 \right)=\dfrac{25}{{{\log }_{b}}a+1} \\
\end{aligned} \right.a,b,c>1{{\log }_{b}}c>04\left( 1+{{\log }_{a}}b \right)\left( 1+{{\log }_{b}}a \right)=25\Rightarrow {{\log }_{a}}b=\dfrac{1}{4}{{\log }_{b}}a+{{\log }_{a}}c+{{\log }_{c}}b\ge 4+2\sqrt{{{\log }_{a}}c.{{\log }_{c}}b}=4+2\sqrt{{{\log }_{a}}b}=55a={{b}^{4}},a={{c}^{2}},c={{b}^{2}}$.