Convex function inequalities 1) Let $f:\mathbb{R}\to\mathbb{R} $ be a positive, convex, continuous function. Assume $f$ satisifes the following inequality$$f(x)f(y)\leq f(xy)$$
for all $x,y\in \mathbb{R}.$
What can we say most about $f$?
Update:
2) Let $f:[0,+\infty)\to\mathbb{R} $ be a non-negative, convex, continuous function. Assume $f$ satisifes the following inequality$$f(x)f(y)\leq f(xy)$$
for all $x,y\in \mathbb{R}.$
What can we say most about $f$?
 A: In 1), the conditions imply that $f$ is constant.
Since $f$ is positive, we may set $y=0$ in the inequality and conclude that $0 < f \leq 1$. I claim this shows that $f$ is constant. Seeing this requires a small fact: the definition of convex implies that $f$ lies under the secant line on the interval $[a,b]$ for any $a,b$. But we also have that $f$ lies above the secant line on $[b,\infty)$ and $(-\infty,a]$. To show this, suppose there is $c \in [b,\infty)$ such that $f(c)$ lies under the secant line. More precisely, this means that $$f(c) < \frac{f(b) - f(a)}{b-a}c + \frac{af(b) - bf(a)}{b-a}.$$ It is not so hard to rearrange this inequality to obtain $$f(b) > \frac{f(c) - f(a)}{c-a}b + \frac{cf(a)-af(c)}{c-a}.$$ Clearing the fractions helps the computation. But this shows that $f$ is not convex! A similar argument shows that $f$ must lie above the secant line on the other side, as well.
Hence, we now assume $f$ is not constant. Suppose $f(a) < f(b)$ for some $a < b$. But then the secant line has positive slope $\frac{f(b) - f(a)}{b-a}$. Since $(c,f(c))$ must lie above this secant line for all $c \geq b$, $f$ cannot be bounded above by $1$ for large values. The case $f(b) < f(a)$ is similar. 
For 2), this argument implies that $f$ is decreasing. We certainly do not get that $f$ is constant, since something like $f(x) = \frac{1}{x+1}$ satisfies the conditions.
