Proving or disproving a statement about continuity and limits 
So I've spent a few hours trying to find a counter example to this statement, and then gave up and spent additional few hours trying to prove it and Ive finally became frustrated with this problem. :)
I seem to get stuck in the scenario where $f\left(\frac{1}{x}\right)$ < 0, I'm just unable to find an epsilon that will satisfy all the conditions.. plus I'm not even sure if the statement is true or false anymore..
 A: Take the limit in your inequality as $x \to 1$, which gives
$$
\lim_{x \to 1} f(x) f(1/x) = f(1)^2 \leq 0,
$$
where the equality uses the continuity of $f$ at $1$. Since a square is nonnegative, the only possibility that remains is $f(1) = 0$.
A: Let's attempt a formal approach:
Let $f(1) = a \gt 0$.
Then for $\epsilon > 0, \exists \delta > 0: |x-1| < \delta \implies |f(x) - a| < \epsilon$.
Taking $\epsilon = a/2; f(x) \in (a/2, a), \forall x \in (1-\delta,1+\delta) $.
But we can choose $\alpha: \alpha, 1/\alpha \in (1-\delta, 1+\delta)$. Thus $f(\alpha)f(1/\alpha)>0$.
This gives us a contradiction. And hence $ f(1)\ngtr 0$.
Similarly we can show $f(1) \nless 0$.
And hence $f(1)=0$.

Additionally, it can be shown that $1 \lt \alpha \lt \min(1+\delta, \frac 1 {1-\delta})$ suffices above. Here's how:
One way of choosing $\alpha: \alpha, 1/\alpha \in (1-\delta, 1+\delta)$ is,
$1< \alpha < 1+\delta; 1-\delta<1/\alpha<1$ (we take $\alpha>1; \implies 1/\alpha < 1$)
$\implies 1< \alpha < 1+\delta; 1<\alpha<\frac{1}{1-\delta} \implies 1<\alpha < \min(1+\delta, \frac 1 {1-\delta})$
(you can take $\alpha<1$ and find another possible choice for $\alpha$)
A: Take $ (x_n) \to 1 $. Can you find the limit of $ \left( \dfrac{1}{x_n} \right) $. Now assume $ f(1) \ne 0 $, so without loss of generality we can claim $ f(1) > 0 $.
Try reaching a contradiction using $ f(x_n) $ and $ f(\frac{1}{x_n}) $.
