<s>[INST] Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially. 
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? $(x+3)^2 \ge 0$
Am I missing the point in anyway?
 [/INST] Let $\displaystyle y=\frac1{x^2+6x+10}\ \ \ \ (1)$
On rearrangement, $x^2y+6yx+10y-1=0\ \ \ \ (2)$ which is a Quadratic Equation is $x$ 
As $x$ is real, the discriminant must be $\ge0$
$\implies (6y)^2-4y(10y-1)\ge0\iff y^2-y\le0\iff y(y-1)\le 0$
$\implies 0\le y\le1$
But if $y=0,(2)\implies -1=0$  for finite $x$ which is impossible
If $y=1, x^2+6x+9=0\implies (x+3)^2=0$
$\implies 0<y\le1$
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