Upper bound for a+b+c in terms of ab+bc+ac - MathOverflow

Upper bound for a+b+c in terms of ab+bc+ac

2010-04-12T11:39:06Z

I am given a triple of positive integers $a,b,c$ such that $a \geq 1$ and $b,c \geq 2$.

I would like to find an upper bound for $a+b+c$ in terms of $n = ab+bc+ac$. Clearly $a+b+c < ab+bc+ac = n$.

Is there any sharper upper bound that could be obtained (perhaps asimptotically)?

Answer by Andrej Bauer for Upper bound for a+b+c in terms of ab+bc+ac

2010-04-12T12:15:36Z

The bound $n/2 + 1$ is tight. First, it is a bound because

$a + b + c \leq a b/2 + b c /2 + a c \leq n/2 + a c / 2 \leq n/2 + 1$

Equality is achieved when $a = 1$ and $b = c = 2$, since then we get $a + b + c = 5$ and $n/2 + 1 = (2+4+2)/2 + 1 = 5$.

Note that there might be other tight bounds that are also functions of $n$.

Answer by Cap Khoury for Upper bound for a+b+c in terms of ab+bc+ac

2010-04-12T12:18:59Z

This is (mostly) a pretty routine optimization problem. The methods of (for example) a standard calculus class are enough to tell you that $a+b+c$ will be largest when two of $a,b,c$ are as small as possible and the third is whatever it has to be. So if you don't care whether the variables are integers, take $a=1,b=2,c=(n-2)/3$.

Thus if $n$ has the form $3k+2$, the optimum is achieved in integers. Take $a=1,b=2,c=k$, and then we have $a+b+c=3+\frac{n-2}{3}=\frac{n+7}{3}$.

So $\frac{n+7}{3}$ is an upper bound, and it actually gives the correct answer infinitely often.