Question

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

Answer 1

Let $y=\frac{1}{3x+1}+\frac{1}{x+1}-\frac{1}{(3x+1)(x+1)}$

$\Rightarrow y=\frac{(x+1)+(3x+1)}{(3x+1)(x+1)}-\frac{1}{(3x+1)(x+1)}$

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

$\Rightarrow y(3x^2+4x+1)=4x+1$

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

No since $x$ is real,

So the discriminant of above quadratic equation should be no-negative

$\Rightarrow 16(y-1{)}^2-12y(y-1)\ge 0$

$\Rightarrow 4(y-1{)}^2-3y(y-1)\ge 0$

$\Rightarrow (y-1)\left[4\right(y-1)-3y]\ge 0$

$\Rightarrow (y-1)(y-4)\ge 0$

$\Rightarrow y\in (-\infty ,1]\cup [4,\infty )$

That means range of given expression does not lie between $1$ and $4$