Prove that $\frac{1}{3\mathrm{x}+1}+\frac{1}{\mathrm{x}+1}-\frac{1}{(3\mathrm{x}+1)(\mathrm{x}+1)}$ doesn’t lie between 1 and 4, if x is real

$\begin{array}{l}\mathrm{y}=\frac{1}{3\mathrm{x}+1}+\frac{1}{\mathrm{x}+1}-\frac{1}{(3\mathrm{x}+1)(\mathrm{x}+1)}\\ =\frac{\mathrm{x}+1+3\mathrm{x}+1-1}{(3\mathrm{x}+1)(\mathrm{x}+1)};\mathrm{y}=\frac{4\mathrm{x}+1}{(3\mathrm{x}+1)(\mathrm{x}+1)}\\ \mathrm{y}(3\mathrm{x}+1)(\mathrm{x}+1)=4\mathrm{x}+1\\ \mathrm{y}\left(3{\mathrm{x}}^2+4\mathrm{x}+1\right)-4\mathrm{x}-1=0\\ 3{\mathrm{yx}}^2+4\mathrm{yx}+\mathrm{y}-4\mathrm{x}-1=0\\ 3{\mathrm{yx}}^2+4(\mathrm{y}-1)\mathrm{x}+\mathrm{y}-1=0\end{array}$
Compare with ax2 + bx + c = 0
we get a = 3y, b = 4 (y–1), c = y–1
Since x is real, $\mathrm{\Delta }\ge 0$
$\begin{array}{l}\Rightarrow {\mathrm{b}}^2-4\mathrm{ac}\ge 0\Rightarrow \left[4\right(\mathrm{y}-1){]}^2-4(3\mathrm{y}\left)\right(\mathrm{y}-1)\ge 0\\ \Rightarrow 16\left[{\mathrm{y}}^2-2\mathrm{y}+1\right]-4\left[3{\mathrm{y}}^2-3\mathrm{y}\right]\ge 0\\ 4\left[4\left({\mathrm{y}}^2-2\mathrm{y}+1\right)-\left(3{\mathrm{y}}^2-3\mathrm{y}\right)\right]\ge 0\\ \Rightarrow 4{\mathrm{y}}^2-8\mathrm{y}+4-3{\mathrm{y}}^2+3\mathrm{y}\ge 0\\ {\mathrm{y}}^2-5\mathrm{y}+4\ge 0(\mathrm{y}-1)(\mathrm{y}-4)\ge 0\\ \Rightarrow \mathrm{y}\le 1\text{ or }\mathrm{y}\ge 4\end{array}$
y does not lie between 1and 4