Proving the sign of a function

Question

How can we prove that

$$\frac{2}{\sqrt{8 n+4 x^2-12 x+1}}-\frac{(2 x-3) (8 x-12)}{2 \left(8 n+4 x^2-12 x+1\right)^{3/2}}$$

is always positive for $n \geq x \geq 1$?

Answer 1

Assuming that $x$ and $n$ are chosen such that $8n+4x^2-12x+1>0$ (i.e. this expression is defined)...

We have $\sqrt{8n+4x^2-12x+1}>0$ so multiplying by this won't change the sign. Thus we could consider whether $$2-\dfrac{(2 x-3) (8 x-12)}{2 \left(8 n+4 x^2-12 x+1\right)}$$ is positive.

Notice that $4x^2-12x+(8n+1)$ is a quadratic in $x$. It's discriminant is $(-12)^2-4(4)(8n+1)=144-16(8n+1) = 128(1-n)$. $n \geq 1$ so either this discriminant is 0 or negative. Either we have a repeated root or complex roots. Either case implies that $4x^2-12x+(8n+1)$ is always non-positive or always non-negative (it's graph either doesn't cross the $x$-axis or it touches it in one place). Since the leading coefficient is $4$, it must be non-negative (this parabola "opens up"). All that said, we can multiply through by $4x^2-12x+(8n+1)$ and not change the sign (because it's positive).

Now we can ask if $$2(2(8n+4x^2-12x+1))-(2x-3)(8x-12)$$ is positive or not.

But $2(2(8n+4x^2-12x+1))-(2x-3)(8x-12) = 32(n-1)$

So it's positive as long as $n>1$.