# 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

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What are f and g in this case? – Nuno Nov 10 '10 at 14:49
some functions..., and they are greater or equal to 1 – limath Nov 10 '10 at 14:50
Of course this rather depends on what $f$ and $g$ are. – Robin Chapman Nov 10 '10 at 14:51
is it possible to refute it, cause I need to choose what to do with it prove or refute? – limath Nov 10 '10 at 14:52

Hint: $2.25 < 3$.

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Just to be annoying: plot $x^2$ and its derivative on the same axes. – J. M. Nov 10 '10 at 15:03
I think that limath just wants f(n) * g(n) $\geq$ f(n) + g(n) eventually. In other words, for some $N \geq 1$, then f(n) * g(n) $\geq$ f(n) + g(n) for all $n \geq N$. – Gabe Cunningham Nov 10 '10 at 15:27
It just occurred to me that you might be using constant functions instead of f(x) = x^2 and g(x) = x. – Gabe Cunningham Nov 10 '10 at 15:32
@Gabe: Yes I meant constant. – Aryabhata Nov 10 '10 at 16:03

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.

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If I may ask, how did you come across this question over a year after it was last touched? – mixedmath Jun 8 '12 at 12:58

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, $1 \cdot x < 1+x$.

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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?

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Not sure what you are getting at. f(n) = g(n) = 3/2 + 1/n. – Aryabhata Nov 10 '10 at 16:02
Oops, I guess I was implicitly assuming that the derivatives were positive. – Gabe Cunningham Nov 10 '10 at 18:17
Even if the derivatives were positive: f(n) = g(n) = 3/2 - 1/10n. – Aryabhata Nov 11 '10 at 19:36