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?
 A: Since $(x+3)^2\geq 0$ for all real values of $x$, $x^2+6x+10=x^2+6x+9+1=(x+3)^2+1\geq 1$, and the inequality follows.
A: Does this parabola (in the denom) open up or down? What is the y-coordinate of its vertex?
A: We are given $\frac{1}{x^2 + 6 x + 10}$ which we can easily rearrange to $\frac{1}{(x+3)^2 +1}$.
If we consider the maximum value of $(x+3)^2 + 1$ then clearly for large positive or negative numbers this can become arbitrarily large and will approach $+\infty$  for large positive or negative $x$.  We are dividing 1 by a potentially large positive number so the result will get close to but will always be more than zero.
We have thus shown:
$$0 \lt \frac{1}{x^2 + 6 x + 10}$$
Now consider the minimum value of $(x+3)^2 + 1$, Since $(x+3)^2$ can not be negative for real $x$ the minimum is 1 at $x = -3$. And 1 divided by 1 is 1 so we have proved the other side of this inequality
$$0 \lt \frac{1}{x^2 + 6 x + 10} \le 1$$
And the resultant function will look like this:

A: 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 $y=1, x^2+6x+9=0\implies (x+3)^2=0$
$\implies 0<y\le1$
