Cho...

Không mất tính tổng quát, giả sử $x\ge y\ge z\Rightarrow \min \left\{ x,y,z \right\}=z$
Ta có: $M=\dfrac{9}{{{x}^{2}}+3z-x+1}+\dfrac{9}{{{y}^{2}}+3x-y+1}+\dfrac{9}{{{z}^{2}}+3y-z+1}-\dfrac{1}{2z-3}$
Do $y\ge z$ nên ${{z}^{2}}+3y-z+1={{z}^{2}}-z+1+3y\ge {{z}^{2}}-z+1+3z={{\left( z+1 \right)}^{2}}$
Suy ra: $\dfrac{9}{{{z}^{2}}+3y+1-z}\le \dfrac{9}{{{\left( z+1 \right)}^{2}}},\forall z\ge \dfrac{7}{4} (*)$

Xét hàm số $f\left( t \right)={{t}^{2}}-t$ với $t\ge \dfrac{7}{4}$. Ta có ${f}'\left( t \right)=2t-1>0,\forall t\ge \dfrac{7}{4}$
Bảng biến thiên:
Dựa vào bảng biến thiên, ta có
$x\ge y\ge z\ge \dfrac{7}{4}\Leftrightarrow f\left( x \right)\ge f\left( y \right)\ge f\left( z \right)\Rightarrow \left\{ \begin{aligned}
& f\left( x \right)\ge f\left( z \right) \\
& f\left( y \right)\ge f\left( z \right) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-x\ge {{z}^{2}}-z \\
& {{y}^{2}}-y\ge {{z}^{2}}-z \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-x+3z+1\ge {{z}^{2}}-z+3z+1 \\
& {{y}^{2}}-y+3x+1\ge {{z}^{2}}-z+3x+1 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+3z-x+1\ge {{z}^{2}}+2z+1={{\left( z+1 \right)}^{2}} \\
& {{y}^{2}}+3x-y+1\ge {{z}^{2}}-z+3x+1\ge {{z}^{2}}-z+3z+1 (do x\ge z) \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+3z-x+1\ge {{\left( z+1 \right)}^{2}} \\
& {{y}^{2}}+3x-y+1\ge {{\left( z+1 \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{9}{{{x}^{2}}+3z-x+1}\le \dfrac{9}{{{\left( z+1 \right)}^{2}}} \\
& \dfrac{9}{{{y}^{2}}+3x-y+1}\le \dfrac{9}{{{\left( z+1 \right)}^{2}}} \\
\end{aligned} \right. (**)$
Từ (*) và (**), ta có $M\le \dfrac{9}{{{\left( z+1 \right)}^{2}}}+\dfrac{9}{{{\left( z+1 \right)}^{2}}}+\dfrac{9}{{{\left( z+1 \right)}^{2}}}-\dfrac{1}{2z-3}$
$\Leftrightarrow M\le \dfrac{27}{{{\left( z+1 \right)}^{2}}}-\dfrac{1}{2z-3}=\dfrac{27}{{{\left( z+1 \right)}^{2}}}-\dfrac{1}{2\left( z+1 \right)-5}=\dfrac{27}{{{u}^{2}}}-\dfrac{1}{2u-5}=g\left( u \right)$, với $u=z+1\ge \dfrac{11}{4}$
Xét hàm $g\left( u \right)=\dfrac{27}{{{u}^{2}}}-\dfrac{1}{2u-5}$ với $u\ge \dfrac{11}{4}$. Ta có:
${g}'\left( u \right)=-\dfrac{54}{{{u}^{3}}}+\dfrac{2}{{{\left( 2u-5 \right)}^{2}}}=\dfrac{2\left( {{u}^{3}}-108{{u}^{2}}+540u-675 \right)}{{{u}^{3}}{{\left( 2u-5 \right)}^{2}}}$
${g}'\left( u \right)=0\Leftrightarrow \left[ \begin{aligned}
& u=2,188<\dfrac{11}{4} \\
& u=3 \\
& u=102,8 \\
\end{aligned} \right.$
Bảng biến thiên:

Từ BBT ta suy ra $M\le 2$
$\Rightarrow M\le 3$