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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?

Ali Caglayan
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salman
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    No, that's the point. $(x+3)^2 \geqslant 0 \Rightarrow (x+3)^2 +1 \geqslant 1 \Rightarrow 0 < \frac{1}{(x+3)^2+1} \leqslant 1$. – Daniel Fischer Sep 01 '13 at 15:33
  • @copper.hat $y \gt 0$ (deleted post?) – salman Sep 01 '13 at 15:35
  • It is best not to think immediately in terms of "algebra." I want to show that $\frac{1}{x^2+6x+10}$ is positive but kind of small. So want to show that $x^2+6x+10$ is positive and kind of big. – André Nicolas Sep 01 '13 at 15:39
  • @salman: My deleted post was just commenting that $0 < \frac{1}{y} \le 1$ iff $1 \le y$. (Which is essentially what André wrote.) – copper.hat Sep 01 '13 at 15:42

4 Answers4

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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.

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Does this parabola (in the denom) open up or down? What is the y-coordinate of its vertex?

imranfat
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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: enter image description here

Warren Hill
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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 $y=1, x^2+6x+9=0\implies (x+3)^2=0$

$\implies 0<y\le1$

lab bhattacharjee
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  • why does this method for forming a quadratic in y work? – salman Sep 02 '13 at 14:44
  • @salman, as $y$ must assume such values for which $x$ is real, After all we are dealing with Real Analysis. http://math.stackexchange.com/questions/426437/finding-the-range-of-fx-1-x-1x-2, http://math.stackexchange.com/questions/443322/what-is-the-maximum-value-of-frac2xx-1-fracxx-1-if-x-in-ma, http://math.stackexchange.com/questions/472169/find-extreme-values-of-frac2xx4, http://math.stackexchange.com/questions/264504/finding-range-of-fracx1-x2 – lab bhattacharjee Sep 02 '13 at 18:06