Maxima of a given set of integers

Question

If $a,b,c$ and $d$ are four different positive integers selected from $1$ to $25$, then what is the highest possible value of the following?$$\frac{a+b+c+d}{a+b+c-d}$$

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I assumed $a+b+c =x$ and then :-

$\frac{(a+b+c+d)}{(a+b+c-d)} \\ \Rightarrow \frac{x+d}{x-d} \\ \Rightarrow \frac{1+d/x}{1-d/x} \\ \Rightarrow Assuming \ d/x \ to \ be \ 'r' \ hence \frac{1+r}{1-r}$

From here I am kind of lost and couldn't figure out a way to proceed also using derivatives didn't help much. Can someone please help me on this ?

Update : If I take $d = 25$ in order to maximize the numerator and minimize the denominator then we'll get $\frac{x+25}{x-25}$.

Thanks in advance !

Answer 1

$\frac {x+d}{x-d} = \frac {(x-d) + 2d}{x-d} = 1 + \frac {2d}{x-d}$.

Now if $x-d = 0$ the whole thing is undefined. Otherwise $x-d \ge 1$ or $x-d < 0$ and $0< d\le 25$.

If $x-d < 0$ then as $d > 0$ we have $1 + \frac {2d}{x-d} < 1$.

If $x-d \ge 1$ then

So $1 +\frac {2d}{x-d} \le 1 + \frac{2d}1 \le 1 + \frac {2\cdot 25}1 = 51$ is the absolute maximum.

Which can be achieved if $d=25$ and ${x-d} = a+b+c-d = a+b+c - 25 = 1$ or if $a+b+c = 26$

(Obviously $a+b+c = 26$ is possible. We can let $a=1; b= 2$ and $c = 23$ or just about any other combination. The requirement that $a,b,c,d$ be distinct my put limits on generalizing solutions [It'd be irritating to prove abstractly that $1 \le a \le 24; 1\le b \le 25- a;b\ne a; c= 26-a-b; c\ne b;c\ne a$ has solutions] but this is so general and easy that it isn't a serious burden... for this problem.)

Answer 2

What you've done so far is reasonable. Derivatives won't help, because the variables are required to be integers. Since $\lim_{r\to1}\frac{1+r}{1-r}=\infty$, to get a large value, we want to make $r$ close to $1$, or $a+b+c$ close to $d$. At the same time, we want to make $a+b+c+d$ as large as we can.

Let's try $a+b+c-d=1$, which is as small as we can make the denominator, and still have a positive value. Then the numerator is $2d+1$ which is largest if we make $d=25$. We can take $a,b,c$ to be any distinct positive integers such that $a+b+c=26$, say $a=7,\ b=9,\ c=10.$ This gives the largest fraction, $51$.

If we allow the denominator to be a little bigger in the hopes of getting a larger, we quickly see that it doesn't work. If $a+b+c-d=2$, then $a+b+c+d =2d+2$, and even with $d=25$, we don't get more than $26$.