I've solved part a. And for part b, I solved $g'\left(x\right)^2$ and when c=0 or $x=-b/c$, we have the minimum, but according to the problem, we can't reach it. So the minimum occurs when $x^2=2|c|$ and $b^2=9/2|c|$. To finish the proof, I need to find the relation between x and b. And use b to represent x, can I combine these two equations: $x^2=2|c|$ and $b^2=9/2|c|$ and say that $x=2b/3$? And still I don't know how to approach part c, any hints? Thanks a lot!
$$g^2(x)=b^2+\frac{2bc}{x}+\frac{c^2}{x^2}$$
Let $y=\frac{1}{x}$. Then $$g^2(x)=b^2+2bcy+c^2y^2$$ We can check easily that this is a parabola that opens upward and touches the $x$-axis at only one point $y=-\frac{b}{c}$.
Now we want to show that $g^2(x)\geq 2|c|$ if $x^2\geq 2|c|$, i.e., $$-\frac{1}{\sqrt{2|c|}}\leq y \leq \frac{1}{\sqrt{2|c|}}$$ for some region of $b$.
There are two cases:
You can see from the picture, both cases should work. We need to find the intersection of the line $y=\frac{1}{\sqrt{2|c|}}$ and the parabola. That point is where $g^2(x)=2|c|$. When $-\frac{b}{c}>0$, it must also satisfy $-\frac{b}{c}>\frac{1}{\sqrt{2||c}}$. When $-\frac{b}{c}<0$, it must also satisfy $-\frac{b}{c}<-\frac{1}{\sqrt{2||c}}$.
Now there are four cases to discuss:
In both working cases, when $b^2\geq \frac{9}{2}|c|$, they still satisfy the criteria for $y$.
For part (c), you just need to draw the curve $b^2=\frac{9}{2}|c|$, shade the region where $b^2\geq\frac{9}{2}|c|$ on the $bc$-plane. It should be a parabola passing through the origin. The convergence comes from part (a) and (b). We see that whenever $x_0^2\geq 2|c|$, by part (b), all the following points $x_n^2\geq 2|c|$. This guarantees that $|g'(x)|\leq \frac{1}{2}$. This also gives the convergence rate.
