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.

Combine the expression into a single rational function $y(x)$, then use the real-discriminant condition on the resulting quadratic in $x$ to bound the values $y$ can take.

Let $y = \dfrac{1}{3x+1}+\dfrac{1}{x+1}-\dfrac{1}{(3x+1)(x+1)}$. Combining over the common denominator $(3x+1)(x+1)$:

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

For a real $x$ to exist and produce a given value $y$, rearrange as a quadratic in $x$:

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

Since $x$ is real, the discriminant of this quadratic (in $x$) must be $\ge0$:

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

$16(y-1)^2-12y(y-1)\ge0 \ \Rightarrow\ 4(y-1)\big[4(y-1)-3y\big]\ge0 \ \Rightarrow\ 4(y-1)(y-4)\ge0.$

So $(y-1)(y-4)\ge0$, which means

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