Cho $x,y$ là các số thực dương thỏa mãn ${{\log...

Ta có ${{\log }_{3}}\dfrac{2x+y+1}{x+y}=x+2y$
$\Leftrightarrow {{\log }_{3}}\left( 2x+y+1 \right)-{{\log }_{3}}\left( x+y \right)=x+2y$
$\Leftrightarrow {{\log }_{3}}\left( 2x+y+1 \right)={{\log }_{3}}\left( 3x+3y \right)+x+2y-1$
$\Leftrightarrow {{\log }_{3}}\left( 2x+y+1 \right)+2x+y+1={{\log }_{3}}\left( 3x+3y \right)+3x+3y\Leftrightarrow f\left( 2x+y+1 \right)=f\left( 3x+3y \right).$
Xét hàm số $f\left( t \right)={{\log }_{3}}t+t$, với $t>0$ ta có ${f}'\left( t \right)=\dfrac{1}{t\ln 3}+1>0,\forall t>0\Rightarrow f\left( t \right)$ đồng biến trên $\left( 0;+\infty \right)$
$\Rightarrow 2x+y+1=3x+3y\Leftrightarrow x+2y=1\Leftrightarrow x=1-2y\Rightarrow T=\dfrac{1}{x}+\dfrac{2}{\sqrt{y}}=\dfrac{1}{1-2y}+\dfrac{2}{\sqrt{y}}.$
Vì $x,y>0\Rightarrow 0<y<\dfrac{1}{2}$. Xét hàm số $g\left( y \right)=\dfrac{1}{2y-1}+\dfrac{2}{\sqrt{y}}$, với $y\in \left( 0;\dfrac{1}{2} \right)$ ta có
${g}'\left( y \right)=\dfrac{2}{{{\left( 2y-1 \right)}^{2}}}-\dfrac{2}{y}.\dfrac{1}{2\sqrt{y}}=0\Rightarrow {{\left( 2y-1 \right)}^{4}}=4{{y}^{3}}\Leftrightarrow 2{{\left( 2y-1 \right)}^{4}}={{\left( 2y \right)}^{3}}.$
Đặt $u=2y-1\in \left( -1;0 \right)\Rightarrow 2{{u}^{4}}={{\left( u+1 \right)}^{3}}\Leftrightarrow 2{{u}^{4}}-{{u}^{3}}-3{{u}^{2}}-3u-1=0$
$\Leftrightarrow {{u}^{3}}\left( 2u+1 \right)-{{u}^{2}}\left( 2u+1 \right)-u\left( 2u+1 \right)-\left( 2u+1 \right)=0$
$\Leftrightarrow \left( 2u+1 \right)\left( {{u}^{3}}-{{u}^{2}}-u-1 \right)=0.$
Với $-1<u<0\Rightarrow {{u}^{3}}-{{u}^{2}}-u-1={{u}^{3}}-{{\left( u+\dfrac{1}{2} \right)}^{2}}-\dfrac{3}{4}<0+0-\dfrac{3}{4}<0\Rightarrow u=-\dfrac{1}{2}\Rightarrow y=\dfrac{1}{4}.$
Từ đó $g\left( y \right)\ge g\left( \dfrac{1}{4} \right)=6\Rightarrow T\ge 6,$ dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& y=\dfrac{1}{4} \\
& x=\dfrac{1}{2} \\
\end{aligned} \right.$.