证法一：记\[ g\left(x\right)=\ln x+{\sqrt{x}}-1-{\dfrac{3}{2}}\left(x-1\right), \]则当 $ x>1 $ 时，\[ g′\left(x\right)={\dfrac{1}{x}}+{\dfrac{1}{2{\sqrt{x}}}}-{\dfrac{3}{2}}<0.\]又 $g\left(1\right)=0$，所以当 $x>1$ 时，有 $ g\left(x\right)<0 $，即\[ f\left(x\right)<{\dfrac{3}{2}}\left(x-1\right). \]证法二：由均值不等式，当 $ x>1 $ 时，$ 2{\sqrt{x}}<x+1 $，故\[ {\sqrt{x}}<{\dfrac{x}{2}}+{\dfrac{1}{2}} \quad \cdots \cdots ① \]令 $ k\left(x\right)=\ln x-x+1 $，则\[ \begin{split}k\left(1\right)&=0,\\k′\left(x\right)&={\dfrac{1}{x}}-1<0,\end{split} \]故 $x>1$ 时，$ k\left(x\right)<0 $，即\[ \ln x<x-1 \quad \cdots \cdots ② \]由 ①② 得，当 $ x>1 $ 时，$ f\left(x\right)<{\dfrac{3}{2}}\left(x-1\right) $．

证法一：记\[ h\left(x\right)=\left(x+5\right)f\left(x\right)-9\left(x-1\right), \]则当 $ 1<x<3 $ 时，由（1）得\[ \begin{split}h′\left(x\right) &=f\left(x\right)+\left(x+5\right)f ′\left(x\right)-9\\&<\dfrac{3}{2}\left(x-1\right)+\left(x+5\right) \left(\dfrac{1}{x}+\dfrac{1}{2{\sqrt{x}}} \right)-9\\&=\dfrac{1}{2x}\left[3x\left(x-1\right)+\left(x+5\right)\left(2+{\sqrt{x}}\right)-18x\right]\\&<\dfrac{1}{2x} \left[3x\left(x-1\right)+\left(x+5\right) \left(2+\dfrac{x}{2}+\dfrac{1}{2} \right)-18x\right] \\&=\dfrac{1}{4x}\left(7x^2-32x+25\right)<0.\end{split}\]因此 $ h\left(x\right) $ 在 $ \left(1,3\right) $ 内单调递减，又 $ h\left(1\right)=0 $，所以 $ h\left(x\right)<0 $，即\[ f\left(x\right)<{\dfrac{9\left(x-1\right)}{x+5}} .\]证法二：记\[ h\left(x\right)=f\left(x\right)-{\dfrac{9\left(x-1\right)}{x+5}} ,\]由（1）得\[\begin{split} h′\left(x\right) &=\dfrac{1}{x}+\dfrac{1}{2{\sqrt{x}}}-\dfrac{54}{\left(x+5\right)^2}\\&=\dfrac{2+\sqrt x}{2x}-\dfrac{54}{\left(x+5\right)^2}\\&<\dfrac{x+5}{4x}-\dfrac{54}{\left(x+5\right)^2}\\&=\dfrac{\left(x+5\right)^3-216x}{4x\left(x+5\right)^2}. \end{split}\]令 $ g\left(x\right)=\left(x+5\right)^3-216x $，则当 $ 1<x<3 $ 时，\[g′\left(x\right)=3\left(x+5\right)^2-216<0 .\]因此 $ g\left(x\right) $ 在 $ \left(1,3\right) $ 内单调递减，又由 $ g\left(1\right)=0 $，得 $ g\left(x\right)<0 $，所以 $ h′\left(x\right)<0 $．
因此 $ h\left(x\right) $ 在 $ \left(1,3\right) $ 内单调递减，又 $ h\left(1\right)=0 $，得 $ h\left(x\right)<0 $．
于是当 $ 1<x<3 $ 时，\[ f\left(x\right)<{\dfrac{9\left(x-1\right)}{x+5}} .\]