Q.Prove that $\dfrac{1}{3x + 1} + \dfrac{1}{x + 1} - \dfrac{1}{(3x + 1)(x + 1)}$ does not lie between $1$ and $4$, if $x$ is real.

After simplifying and using the discriminant condition, $(y-1)(y-4)\ge0$, so the value cannot lie strictly between $1$ and $4$.

Combine over the common denominator $(3x+1)(x+1)$:

$\dfrac{(x+1)+(3x+1)-1}{(3x+1)(x+1)}=\dfrac{4x+1}{3x^2+4x+1}=y \text{ (say)}.$

Then $y(3x^2+4x+1)=4x+1$, i.e.

$3y\,x^2+(4y-4)x+(y-1)=0.$

For $x$ real (and $y e0$) the discriminant must be non-negative:

$(4y-4)^2-4(3y)(y-1)\ge0$

$16(y-1)^2-12y(y-1)\ge0$ …