# Inequality

• Nov 11th 2009, 02:09 AM
Showcase_22
Inequality
Prove that $\forall \ x \in \mathbb{R}, \ \left |\frac{x}{n(1+nx^2)} \right| \leq \frac{1}{2n^{\frac{3}{2}}}$

Everything I try isn't working that easily. I start by doing $\frac{x}{n(1+nx^2)} \leq \frac{x}{n^2(1+x^2)}$ but i'm not really sure what to do from here!
• Nov 11th 2009, 03:31 AM
tonio
Quote:

Originally Posted by Showcase_22
Prove that $\forall \ x \in \mathbb{R}, \ \left |\frac{x}{n(1+nx^2)} \right| \leq \frac{1}{2n^{\frac{3}{2}}}$

Everything I try isn't working that easily. I start by doing $\frac{x}{n(1+nx^2)} \leq \frac{x}{n^2(1+x^2)}$ but i'm not really sure what to do from here!

$\left |\frac{x}{n(1+nx^2)} \right| \leq \frac{1}{2n^{\frac{3}{2}}}\Longleftrightarrow |x|2n^{3\slash 2}\leq n(1+nx^2)$ $\Longleftrightarrow n^2x^2-2n^{3\slash 2}|x|+n\geq 0$

Now, the last expression from the right is a quadratic in x, and it never minds whether we have $x\geq 0\,\,\,or\,\,\,x < 0$ , its discriminant it's the same. Well, evaluate its discriminant and solve the problem.

Tonio
• Nov 11th 2009, 03:42 AM
Showcase_22
I thought I could get there via an inequality route. =S

Discriminant= $4n^3-4n^2(n)=4n^3-4n^3=0$

This implies that the quadratic only has 1 root and this root is repeated. Therefore a graph of this quadratic would have a minimum point that touches the x axis. Hence the entire quadratic would be greater than 0.

For a proof though, would it be better to construct the quadratic first, work all this out and eventually arrive at it must be greater than 0?
• Nov 11th 2009, 04:00 AM
tonio
Quote:

Originally Posted by Showcase_22
I thought I could get there via an inequality route. =S

Discriminant= $4n^3-4n^2(n)=4n^3-4n^3=0$

This implies that the quadratic only has 1 root and this root is repeated. Therefore a graph of this quadratic would have a minimum point that touches the x axis. Hence the entire quadratic would be greater than 0.

For a proof though, would it be better to construct the quadratic first, work all this out and eventually arrive at it must be greater than 0?

Perhaps, though it usually is JHS or at most HS stuff to show that if the discriminant vanishes then the quadratic is a perfect square.
Anyway, in your case $n^2x^2-2n^{3\slash 2}|x|+n=\left(nx\pm \sqrt{n}\right)^2$ , the sign being minus or plus according as whether $x\geq \,\,or\,\, x < 0$ , resp.

Tonio
• Nov 11th 2009, 11:17 AM
Showcase_22
JHS and HS?? =S

Thanks for your help, I think this question is pretty much done.