The convex combination of two unequal numbers is strictly between them $ x,y,a\; \in F $ which is an ordered field and this is given:
$$ 0 \lt a \lt 1 $$
$$ x \lt y $$
I need to prove:
$$ x\lt ax + (1-a)y\lt y $$
This is a relatively simple proof, yet I'm having quite the difficulty with it. What's the strategy for attacking proofs? Brute forcing it (i.e. trying many different steps without too much thought isn't working) isn't working well for me. 
I've tried starting with $ 0 \lt a \lt 1 $ and doing ordered field operations (adding to each side, multiplying by each side etc) but I'm unable to get anywhere close to an actual proof.
So I've done this:
$$ 0\lt a \lt 1$$
Multiply sides by (x-y):
$$ 0*(x-y)\lt a(x-y) \lt (x-y)*1 $$
$$ 0\lt a(x-y) \lt (x-y)$$
Adding y:
$$ y+ 0\lt a(x-y)+y \lt x-y+y $$
$$ y\lt a(x-y)+y \lt x $$
Here is where I appear to have difficulties continuing.
 A: There's no step-by-step algorithm for proofs in general,
as far as I'm aware;
even if it's limited to "proofs I am likely to be asked for
on the next exam" I suspect it would be too difficult to
devise (and learn) a method that is guaranteed to work in every case.
It may be possible to gain a little more facility in what
you try when approaching a problem, however.
For example, when I saw this problem, just about the first thought
that came to mind was, "There's a single $y$ on the far right-hand side;
what do I get if I cancel it? Can I prove the resulting inequality?"
Indeed, if we subtract $y$ from each side of $ax+(1−a)y<y$,
we get
$$
ax - ay < 0,
$$
which seems easy enough to prove, since $x < y$ and $a > 0$.
Then I started to try the same thing for the $x$ on the far left, but
then I wondered which makes more sense, to cancel the $x$ on the left
of $x < ax + (1-a)y$, or to cancel the $ax$ on the right?
It turns out that if we cancel $x$ we get $0 < (a - 1)x + (1-a)y$,
which is provable, but if we cancel $ax$ we get directly to
$(1-a)x < (1 - a)y$, which seems simpler.
It then occurred to me that maybe it's nicer to cancel $(1-a)y$ rather
than $y$ on the right, and indeed that gets us $ax < ay$.
But whether you chose the "nice" way or the "less nice" way, the
task would then be to prove the new fact and use it to prove the
original fact. The pattern seems to be, for each of the inequalities
in the original statement, you start with $x<y$,
then multiply by either $a$ or $1-a$ (either one is positive, so it
preserves the $<$), possibly collect terms on one side of the $>$, 
then add the necessary quantity, $ax$, $(1-a)y$, $x$, or $y$
(depending on which of the two inequalities you're working on and
the proof path you chose) in order to achieve
the desired inequality.
