Upper bound for a+b+c in terms of ab+bc+ac - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T22:08:03Z http://mathoverflow.net/feeds/question/21088 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21088/upper-bound-for-abc-in-terms-of-abbcac Upper bound for a+b+c in terms of ab+bc+ac Jernej 2010-04-12T11:39:06Z 2011-12-01T19:40:55Z <p>I am given a triple of positive integers $a,b,c$ such that $a \geq 1$ and $b,c \geq 2$.</p> <p>I would like to find an upper bound for $a+b+c$ in terms of $n = ab+bc+ac$. Clearly $a+b+c &lt; ab+bc+ac = n$.</p> <p>Is there any sharper upper bound that could be obtained (perhaps asimptotically)?</p> http://mathoverflow.net/questions/21088/upper-bound-for-abc-in-terms-of-abbcac/21097#21097 Answer by Andrej Bauer for Upper bound for a+b+c in terms of ab+bc+ac Andrej Bauer 2010-04-12T12:15:36Z 2010-04-12T12:15:36Z <p>The bound $n/2 + 1$ is tight. First, it is a bound because</p> <blockquote> <p>$a + b + c \leq a b/2 + b c /2 + a c \leq n/2 + a c / 2 \leq n/2 + 1$</p> </blockquote> <p>Equality is achieved when $a = 1$ and $b = c = 2$, since then we get $a + b + c = 5$ and $n/2 + 1 = (2+4+2)/2 + 1 = 5$.</p> <p>Note that there might be other tight bounds that are also functions of $n$.</p> http://mathoverflow.net/questions/21088/upper-bound-for-abc-in-terms-of-abbcac/21098#21098 Answer by Cap Khoury for Upper bound for a+b+c in terms of ab+bc+ac Cap Khoury 2010-04-12T12:18:59Z 2010-04-12T12:18:59Z <p>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 $a=1,b=2,c=(n-2)/3$.</p> <p>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 $a+b+c=3+\frac{n-2}{3}=\frac{n+7}{3}$.</p> <p>So $\frac{n+7}{3}$ is an upper bound, and it actually gives the correct answer infinitely often.</p>