Real positive mappings

[fantispug]

Homework Statement

Prove that for any
$$f:\mathbb{R}\rightarrow\left(0,\infty \right)$$
there is a rational number q and an irrational number i such that
$$f(q)f(i) > \left( \frac{q-i}{2} \right)^2$$
(Or if this is false find a counterexample)

Homework Equations

The Attempt at a Solution

The best idea to me seems to be to look for a contradiction, that is assume
$$f(q)f(i) \leq \left( \frac{q-i}{2} \right)^2$$
for all rational q and irrational i and try to come up with a contradiction.

The first approach I tried was to fix a q, say q=0, then for every epsilon greater than zero there exists an irrational number i such that
$$f(i)\leq \epsilon$$
(since every neighbourhood of a real number contains a rational number)
similarly there exists a rational number q such that
$$f(q)\leq \epsilon$$
So both these conditions are fine on their own, so I need to think of some way to combine them.

Define a sequence of irrational numbers, i_0(n) by
$$f(i_0(n)) \leq 1/n$$
Then define interlaced sequences of rational and irrational numbers:
$$f(q_k(n)) \leq f(i_{k-1}(n))$$
$$f(i_k(n)) \leq f(q_k(n))$$
But I haven't found anything useful to do with these sequences - they could be quite trivial, e.g.f(q_k(n))=f(i_k(n)) for all k,n and q_k(n)=q_l(n) for all n, l and k similarly for the irrationals.

So I'm stuck, I've got to think of some way to impose both conditions (the one on the rationals and the one of the irrationals) simultaneously (no single sequence will do the trick), but I can't think of a way to do it.

The only other way I can think of trying to approach it is with some abstract formalism (e.g. topology), but since f isn't even continuous topology won't do. Maybe an argument based on cardinality could be feasible, but I can't think of one.

 

[mighty2000]

Thank you for your post. I appreciate your attempt at finding a solution to this problem. I agree that looking for a contradiction is a good approach. However, I believe there is a simpler way to prove this statement.

Consider the function g(x) = (x-i)^2 - 4f(i)f(x). This is a continuous function on the real line, and since f(i) is always positive, we know that g(i) < 0. This means that there exists some interval around i where g(x) < 0, since g is continuous.

Now, since g(q) is a continuous function on the rationals, and the rationals are dense in the reals, we know that there exists a rational number q in this interval where g(q) < 0. This means that (q-i)^2 < 4f(i)f(q).

Thus, we have found a rational number q and an irrational number i such that f(q)f(i) > \left( \frac{q-i}{2} \right)^2. This holds for any choice of f:\mathbb{R}\rightarrow\left(0,\infty \right), so the statement is proven.

I hope this helps. Let me know if you have any questions or if you need further clarification.