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I'm proving a theorem that needs this part.

Say $n=a \times b$, where $a$ and $b$ are both integer. I'm looking for an expression for the minimum integer $a+b$.

Is there an existing theorem about this? If yes, then I'll be done with what I am doing. Thank you.

[Question Edited]

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    $10000$ has $1$ and $2$ as factors, and their sum is $3$, while twice the square root of $10000$ is $10$. Or did you mean something else? – 2017-02-02
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    So how many factors are you adding? – 2017-02-02
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    I mean the factors that will give the minimum sum. Sorry for the misunderstanding. – 2017-02-02
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    I don't think you understood my question. The minimum of sum of factors, if there is no restriction to how many, is just adding $1$, namely $$1=1$$ So could you clarify the number of factors you are adding? – 2017-02-02
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    I think the question Orior wants to ask is this: given positive integers $a$, $b$, let $n=ab$, and show that $a+b\ge\lceil2\sqrt n\rceil$. But this should fall out of the inequality of the arithmetic and geometric means. – 2017-02-02
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    First off, I have yet to be knighted. Second, if you want to be sure I see a comment intended for me, you must put @Gerry in it somewhere. And third, it still seems that what you are asking is not what I thought you were asking. It now seems like you are asking, given $n$, find the minimum of $d+(n/d)$ over all divisors $d$ of $n$. But about all you can say about that is that the minimum will be achieved when $d$ is the largest divisor of $n$ not exceeding $\sqrt n$. If $n$ is prime, the minimum will be $n+1$. If $n$ is a square, it will be $2\sqrt n$. In general, somewhere in between. – 2017-02-03
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    Copy @GerryMyerson. Thank you for giving time to my post. It's a big help already. i have cleared some things on my mind. :) – 2017-02-03
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    @GerryMyerson About the cases of being prime and a perfect square, where can I find references for that? Thanks so much. – 2017-02-03
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    If you know what a prime is, and what a square is, then surely you can work it out on your own. Honestly, it's not hard. – 2017-02-03

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I think your claim is mistaken. Take $n=101$. The minimu sum of at least two factors is $$1+101=102 \neq \text{ceiling of }2\sqrt{101}=22 $$ However, we can say that if $ab=n$, then $$a+b \ge 2 \sqrt{ab}=2\sqrt{n}$$ The problem is that the equality condition, which is $a=b$ can only be achieved if $n$ is a square.

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    Thanks for the response. I think there is wrong with my statement. What I mean is the factors of the number that will give the minimum sum. – 2017-02-02
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    Say of 100, the factors are 1*100, 2*50, 4*25, 5*20 and 10*10. And the minimum sum of the factors is 20 (from 10*10) that is equal to 2*sqrt(100). Sorry for the misunderstanding. – 2017-02-02
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    @Orior I don't think I understand you. For example, see my edit. What if $n=101$. – 2017-02-02
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    @Orior Also, the factors of $100$ are $1,2,4,5,20,25,50,100$ so the minimum sum of factos is $1+2=3$ – 2017-02-02
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    I edit the problem again. Again, sorry for the misunderstanding. – 2017-02-02
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    @Orior Your edit does not answer my question as to what happens if $n=101$. The minimum is $102$, which contradicts your claim. Could you explain why this is not sufficient? – 2017-02-02
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    I see. Ok thanks for the clarification. – 2017-02-02
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    @Orior Does my answer answer your question? – 2017-02-02
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    @Orior Youvaaid that this answer clarified matters. But does this answer the question? – 2017-02-02
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    I'm sorry it took time. So there will be two cases? It the number is prime and the other is if it's not prime? – 2017-02-03
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    @Orior The cases divide into if it is a square (or sufficiently close to a square number) and when it is not. – 2017-02-03
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    @Orior So we want $\lfloor \sqrt{n} \rfloor$ to divide $n$. – 2017-02-03
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    Hmmm. Ok. I see. Thanks much. Will look at my argument again. – 2017-02-03
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    But is there an existing theorem about minimal sum of the factors of a number? The sum must be an integer? – 2017-02-03