# A Bostock And Chandler Stumper

Given that $p = \frac{3(x^2+1)}{2x-1}$, show that $p^2 -3(p+3)\ge 0$ provided $x$ is real.

Well, this doesn’t look very nice. Let’s try and attack it with the sledgehammer first and then see if there are any shortcuts.

$p^2$ is $\frac{9(x^4 + 2x^2 + 1)}{(2x-1)^2}$, and $3(p+3)$… well, it’s a mess of brackets. Let’s expand $p$ first to get $\frac{3x^2 + 3}{2x-1}$, then add 3 – or rather $\frac{6x-3}{2x-1}$ – to it to get $\frac{3x^2 + 6x}{2x-1}$; lastly, we need to treble it to get $\frac{9x^2 + 18x}{2x-1}$, or $\frac{9x(x+2)}{2x-1}$.

So now we’re trying to show that $\frac{9(x^4 + 2x^2+1)}{(2x-1)^2} - \frac{9x(x+2)}{2x-1} \ge 0$. I can’t say that’s super-nice, although there’s a factor of 9 that comes out, and we can combine everything into a single fraction to get a left-hand side of:

$\frac{9}{(2x-1)^2}\left( (x^4 + 2x^2 + 1)-(x(x+2)(2x-1))\right)$

The first factor is non-negative (except at $x=1/2$, where it’s undefined). The second is… well, what is it? The cubic is $2x^3 + 3x^2 -2x$, so the factor becomes $(x^4 -2x^3 - x^2 + 2x + 1)$.

Now, it *transpires* that that’s a perfect square, $(x^2 - x -1)^2$ but I think even the Mathematical Ninja wouldn’t spot that.

*I heard that.*

Of course you did, sensei. There’s an approach that involves differentiating and finding the turning points, but this appears to be from *very* early in the textbook and I suspect there’s a better way.

### Approach number 2

What we’re looking at here is a quadratic in $p$, and I’m going to work by contradiction: assume there’s a value of $x$ that makes the inequality false.

So, under what circumstances would $p^2 - 3p - 9$ be negative? It doesn’t factorise, and the quadratic is zero when $p = \frac{3\pm3\sqrt{5}}{2}$. That means, I’m trying to find a value of $x$ such that $\frac{3-3\sqrt{5}}{2} < p < \frac{3+3\sqrt{5}}{2}$.

So, now I’ll sub in for $p$, and try to solve:

$\frac{3-3\sqrt{5}}{2} < \frac{3(x^2+1)}{2x-1} < \frac{3+3\sqrt{5}}{2}$

I can divide everything by 3 to start with:

$\frac{1-\sqrt{5}}{2} < \frac{x^2+1}{2x-1} < \frac{1+\sqrt{5}}{2}$

Now, let’s think about the shape of the graph of $y=\frac{x^2+1}{2x-1}$. When $x$ is large, it looks like $y=\frac{x}{2}$, possibly offset by a constant I don’t care about. When $x$ is close to $\frac{1}{2}$, it looks like a reciprocal graph – so the shape of the graph must come down from just to the right of the vertical asymptote, hit a turning point and then rise towards the oblique one; to the left of the vertical asymptote, it does something similar. Here’s Desmos.

So: where are the turning points on $y=\frac{x^2+1}{2x-1}$? If we’re not allowed calculus, we can do something else clever: we can say “where does $\frac{x^2 + 1}{2x-1}=k$ have a double root?” And *that’s* something we can solve using discriminants.

If $\frac{x^2 + 1}{2x-1}=k$, then $x^2 +1 = k(2x-1)$, or $x^2 - 2kx + 1 + k = 0$.

That has double roots when its discriminant is zero, so $4k^2 -4(1+k)=0$ – or $4k^2 - 4k - 4= 0$.

And if we complete the square, we get $(2x-1)^2 = 5$, so $x = \frac{1\pm \sqrt{5}}{2}$, exactly the bounds we had! So, the local minimum is at $y=\frac{1 + \sqrt{5}}{2}$ and the local maximum at $y=\frac{1-\sqrt{5}}{2}$.

And that means, there *isn’t* a value of $x$ that gives us a $p$ that makes the original quadratic negative.

Phew! That was a lot more involved than I expected it to be.