Prove or refute $f(n) * g(n) \ge f(n) + g(n)$ I received some homework (calculus), which I can't prove:
$f(n) * g(n) \ge f(n) + g(n)$ for some $n \geq 1$ is always true $(f(n),  g(n) \ge 1)$
I think that this is true, so I need to prove that there is some $n \geq 1$ ($n$ is a natural number) for which this inequality will be true always, thanks in advance for any idea
 A: Hint: $2.25 < 3$.
A: I know its been a while since I got to exercise the old brain matter, but to disprove this equation, all you need to do is find one example which makes it false.  
So you can take any $n$ in which $f(n)$ or $g(n)$ returns $1$.  And take any $n$ in which $f(n)$ or $g(n)$ does not return $1$, say it returns $x$.  Which would make the equation false, $1 \cdot x < 1+x$.
A: I Hope that i dont understood wrong, but it's false:
Proof:
Supose that is true, and for any case this must work, so for f(n)=1 and g(n)=1, that must work so:
f(n)*g(n)>=f(n)+g(n) => 1*1>=1+1 => 1>=2 that is false. 
So contradiction.And its independly of n.
Also, you can imagine all function that f*g=constant and f+g> that this constant like: 
 => f(n)=1/g(n) 
 => f(n)*g(n)>=f(n)+g(n) 
 => 1>=1/g(n)+g(n)  
and how you supose that g,f>=1  so contradiction.
A: If $f$ and $g$ are both constant functions, then it's easy to see that the claim isn't true in general. Let's assume that $g$ is not constant, and since this problem came from a calculus class, let's assume that $f$ and $g$ are differentiable. Compare the derivative of $f(x) \cdot g(x)$ with the derivative of $f(x) + g(x)$. What can you say about the relative size of the derivatives? What does that tell you about the functions?
