How do I prove that something is a suffcient but not necessary condition? I do not think I understand necessary and sufficient conditions very well. A question I have attempted is phrased as follows:
By considering $(x^2+gx+h)(x-k)$, or otherwise, show that $g^2 > 4h$ is a sufficient condition but not a necessary condition for the inequality $(g-k)^2 > 3(h-gk)$.
So far my argument is this:
If $g^2 > 4h$ then we know that there are two distinct turning points, and therefore $f'(x) = 0$ has two distinct solutions and, therefore the discriminant of $f'(x) > 0 $, after some algebraic manipulation leads me to the following statement:
$(g-k)^2 > 3(h-gk)$.
I am aware that I have not fully answered the question, and I am unsure of how to prove that it is sufficient, but not necessary. I wonder if someone could help with incorporating this into my argument.
I'm not exactly sure what you are asking.
Consider the quadratic $(g-k)^2 -3(h-gk) = k^2+gk -3h +g^2$. Then this will be positive for all $k$ iff it is true for the minimising value of $k$.
The minimising value of $k$ is $k=- {g \over 2}$, and the above expression reduces to ${3 \over 4} (g^2-4h)$.
Hence we see that $g^2 > 4h$ iff $(g-k)^2 -3(h-gk) > 0$ for all $k$. This is an equivalence.
However, if we choose $g=0$, $h=1$ $k=2$ then $(g-k)^2 -3(h-gk) > 0$ but $g^2 \le 4h$.
Hence if $g^2 > 4h$ then $(g-k)^2 -3(h-gk) > 0$ for all $k$, but there are $g,h,k$ such that $(g-k)^2 -3(h-gk) > 0$ for all $k$, but $g^2 \not> 4h$.