Not As Easy As Adding

Motivation : Ironically, I had my motivation to this problem when I was convinced there was no elementary method to solving this (see: something that didn't use something particularly complex like Lagrange Multipliers).

The situation arose when I posed this question to my father, who when taking an interest in this problem, did the usual amateur mistake of thinking the maximum possible value of $ab+ac+bd+cd$ to be the sum of the maximum limits, to which I replied this to be impossible as we'd have $ab*cd=1000*500=ac*bd=1000^2$ .

It is at this point I realised that by possibly considering the product $abcd$ , I might be able to solve this problem.

Method : We first consider the following lemma:

Lemma : Given positive real numbers $a,b$ such that their product is $ab=k$ , and with constraints $a\leq m, b\leq n$ with $k \leq mn$ and $m,n\leq k$ , the maximum possible value of $a+b$ is $\max(m,n)+\frac{k}{\max(m,n)}$ .

Proof : Consider positive real numbers $p,q,r,s,t$ where $p<q<r<s<t$ and $pt=qs=r^2=y$ . We have $2r<q+s<p+t$ or equivalently $r+\frac{y}{r}<s+\frac{y}{s}<t+\frac{y}{t}$ , with the latter inequality easy to prove. (Hint: consider only two sides in this inequality at a time, rearrange all the fraction-like terms on one side and the remaining terms on the other.)

We note with our constants above, we can apply analogous logic to out lemma, that we increase the value of $a+b$ by increasing the value of $|a-b|$ within the given conditions. In the stated conditions of the lemma, this is achieved when $\max(a,b)$ is at its highest, or within the conditions when we have either $a,b=\max(m,n)$

Now in our problem, assume the product $abcd=k$ . With $\max(ab)=1000$ , $\max(cd)=500$ , $\max(ac)=1000$ , $\max(bd)=1000$ , applying our lemma, our desired maximum value is achieved with:

$\max(1000,500)+\frac{k}{\max(1000,500)}+\max(1000,1000)+\frac{k}{\max(1000,1000)}$

$=2000+2*\frac{k}{1000}$

We know that $k \leq 500*1000$ , so our maximum desired value occurs when $k=500*1000$ and our desired value becomes:

$2000+2*\frac{500*1000}{1000}=2000+1000=\boxed{3000}$

Note: as an explicit example, one possible case of this maximum can be achieved when $ab=1000,cd=500,ac=500,bd=1000$ . We can then fix $a,b,c,d$ to be any positive real numbers satisfying these equations. (such as, say $(a,b,c,d)=(2,500,250,2)$ )

(This is not a solution)

The gut response of most people is to say that the conditions imply $ab+ac+bd+cd \leq 3500$ . However, we cannot have equality hold throughout. In particular, if $ab=1000, ac=1000, bd=1000$ , then we must have $cd = \frac{abcd}{ab} = 1000$ , which contradicts the given fact that $cd \leq 500$ .

We fact, all we know is that $ab+ac+bd+cd$ is bounded above by 3500. We do not know if that is the best possible, i.e. least upper bound.

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It is easy to find a solution to $ab+ac+bd+cd = 3000$ , like when $a=500, b=2, c = 1, d = 500$ , or when $a = 1000, d = 500, b = 1, c = 1$ . It remains to show that we always have $ab + ac + bd + cd \leq 3000$ .

Hint: Consider cases where $a \geq 2d, 2d \geq a \geq d, d \geq a$ .

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Hint: If $A \leq 1, B \leq 1$ and $AB \leq 1$ , what is the maximum of $A + B$ ?
If $A \leq 1, B \leq 1$ and $AB \leq 0.8$ , what is the maximum of $A + B$ ?

Hence, conclude that $ac + bd \leq 1000 + 500$ .

In this solution i'll not talk about some particular value of $a, b, c$ or $d$ . Well, to maximize that sum we need to maximize the value of its terms. It's important to observe that we cannot decide to maximize saying $ab = ac = bd = 1000$ . Why? Well, suppose $ab = ac = 1000$ . That would imply $b = c$ , so if you take some value of $d$ such that $bd = 1000$ (with the purpose of maximizing), then as $b = c$ , $bd = cd$ , and $cd \leq 500$ .

So, we can maximize $ab = ac = 1000$ , but not $bd = 1000$ . We'll choose some $d$ such that $cd = 500$ and therefore $bd$ will yield $500$ too. So the maximum of $ab + ac + bd + cd = 1000 + 1000 + 500 + 500 = \boxed{3000}$ .

Lets write the expression

$ab \le 1000, ac \le 1000, bd \le 1000, cd \le 500$

if generally, $cd \le 1000$ we could have simply concluded $a=b=c=d$ , but because it isn't, we can easily see that there is one value which is different from rest and it should either be c or d

Moreover, if the rest of the values are equal then either of c or d is $\frac{1}{2}$ of a or b

Let $d = \frac{a}{2}$ and $a=b=c$

We can the write

$ab+ac+bd+cd = 1000+1000+\sqrt{1000}\frac{\sqrt{1000}}{2}+\sqrt{1000}\frac{\sqrt{1000}}{2}$ $=2000+2\sqrt{1000}\frac{\sqrt{1000}}{2}=2000+\sqrt{1000}^2=3000$

Note

The above solution assumes that $a=b=c$ , but in case $a \ne b \ne c$

We can start with the following expression again

$ab \le 1000$

$ac \le 1000$

$bd \le 1000$

$cd \le 1000$

Then we can safely assume without loss of generality $ab+ac+bd+cd \le 4000$

Let again $d' = \frac{d}{2}$ , then we have

$ab \le 1000$

$ac \le 1000$

$bd' = \frac{bd}{2} \le \frac{1000}{2} \le 500$

$cd' = \frac{cd}{2} \le \frac{1000}{2} \le 500$

So we have $ab+ac+bd'+cd' \le 1000 + 1000 + 500 + 500 = 3000$

This can be extended to any arbitrary sub-multiplier, i.e.

if $d' = \frac{d}{n}$ , then we have

$ab \le 1000$

$ac \le 1000$

$bd' = \frac{bd}{n} \le \frac{1000}{n}$

$cd' = \frac{cd}{n} \le \frac{1000}{n}$

And so we have $ab+ac+bd'+cd' \le 1000 + 1000 + \frac{1000}{n} + \frac{1000}{n} = 2000 + \frac{2000}{n}$

Note , this also proves that, the inequalities are misleading and the original inequalities should had been

$ab \le 1000$

$ac \le 1000$

$bd \le 500$

$cd \le 500$

Q.E.D

Please leave your comments if you feel any flaw in the proof

From given condition, ab+ac =a(b+c)-max=2000. . bd+cd=d(b+c)-max=1500........
Dividing the to equations, a/b=4/3. ~~So if b and c appearing in both equations are maximum possible and a = 4 so that d=3, we have b=c=250. However this makes cd=750. But cd=500. This can be possible if d=2. Since we can have d(b+c)<1500.
So max=4*(250+250)+2(250+250)=3000..
$Since ~each~term~is~a~product~of~two~variables,~and~we~are~working~with~integers~that~has~suitable~factors,\\ WLOG~we~ can~ take~all~variables~as~integers.\\ Let~S~be~the~~sum.~~So~S=ab+ac+db+cd=a(b+c)+d(b+c)....(1).\\ Since~the~limiting~condition~is~cd \leq 500,. ~and~ we ~see~c~appears~in~both~the~factors,\\ it~is ~reasonable~to~assume~for~S_{max}, ~c~must~be~greatest.~\implies~c=500,~and~d=1.\\ \therefore~bd=1000,~\implies~~for~S_{max},~b=1000.~Again~~~ab=1000,~\implies~a=1.\\ \therefore~S_{max}=1000+500+1000+500=3000.\\ Note:-~~There~are~many~combinations~that ~gives~the~same~value.\\ e.g.~(2n,\dfrac{1000}{2n}\dfrac{1000}{2n},n)~~~or~~(n,\dfrac{1000}{2n}\dfrac{500}{n},n),~~~~where~n=1,2,4,5,25,125.$

$Let\quad us\quad understand\quad the\quad given\quad conditions\quad graphically.\quad \\ If\quad a,b,c,d\quad are\quad Real\quad nos.\quad then\quad each\quad \quad inequality\quad independantly\\ represents\quad a\quad set\quad of\quad hyperbolic\quad functions,\quad whose\quad values\quad are\quad \\ satisfied\quad with\quad area\quad between\quad the\quad +ve\quad X\quad axis\quad and\quad the\quad curve.\quad \\ To\quad find\quad the\quad maximum\quad value,\quad first\quad we\quad would\quad have\quad to\quad find\quad the\quad "region"\quad \\ where\quad all\quad the\quad areas\quad for\quad the\quad given\quad hyperbolas\quad satisfy\quad the\quad given\quad inequalities.\\ It\quad can\quad be\quad seen\quad that\quad cd\le 500\quad will\quad be\quad the\quad hyperbola\quad which\quad is\quad below\quad all\quad the\quad rest\quad hyperbolas.\quad \\ So\quad the\quad obvious\quad choice\quad which\quad will\quad satisfy\quad all\quad the\quad rest\quad of\quad inequalities\quad \\ is\quad to\quad start\quad with\quad cd=500\quad which\quad is\quad the\quad maximum\quad value\quad for\quad the\quad cd\quad and\quad \\ where\quad c,d\quad lie\quad on\quad the\quad curve\\ \\ c=\frac { 500 }{ d } .\\ \\ Now,\quad in\quad above\quad function\quad c\quad is\quad dependent\quad variable\quad and\quad d\quad is\quad independent\quad variable.\quad \\ So\quad consider\quad the\quad inequaltiy\quad b\le \frac { 1000 }{ d } \quad which\quad is\quad actually\quad the\quad area\quad under\quad the\quad hyperbola\quad b=\frac { 1000 }{ d } \\ this\quad hyperbola\quad is\quad above\quad that\quad of\quad c=\frac { 500 }{ d } .\quad So\quad the\quad region\quad where\quad the\quad inequality\quad will\quad be\quad satisfied\quad will\\ be\quad the\quad intersection\quad of\quad areas\quad of\quad both\quad hyperbolas\quad which\quad is\quad in\quad fact\quad c=500/d.\quad So\quad thus\quad the\quad maximum\quad value\quad \\ of\quad bd\le \frac { 1000 }{ d } \quad is\quad bd=500.\\ Now,\quad Dividing\quad cd=500\quad and\quad bd=500\quad we\quad get\\ b=c.\\ So\quad thus\quad the\quad remaining\quad inequalities\quad are\quad b\le \frac { 1000 }{ a } \quad and\quad b\le \frac { 1000 }{ a } .\\ Now\quad The\quad values\quad of\quad b\quad will\quad lie\quad on\quad the\quad hyperbola\quad b=\frac { 500 }{ d } \quad so\quad irrespective\quad of\quad value\quad of\quad b\quad \\ the\quad value\quad of\quad a\quad can\quad so\quad be\quad selected\quad such\quad that\quad their\quad product\quad ab=1000\quad \\ which\quad is\quad the\quad maximum\quad value\quad of\quad ab.\\ \\ So\quad the\quad ans\quad is\quad max\quad value\quad of\quad ab+ac+bd+cd=1000+1000+500+500=3000.$

Ps: Sorry i couldn't attach the graphs as i couldn't find the option. Feel free to ask me any doubts regarding my solution. :)