# find the maximum

• Jan 7th 2008, 11:55 PM
perash
find the maximum
let $a,b,c,d$ be distinct real numbers such that

$a/b+b/c+c/d+d/a=4$

and $ac=bd$
find the maximum value of

$a/c+b/d+c/a+d/b$
• Jan 8th 2008, 01:16 AM
Isomorphism
Quote:

Originally Posted by perash
let $a,b,c,d$ be distinct real numbers such that

$a/b+b/c+c/d+d/a=4$

and $ac=bd$
find the maximum value of

$a/c+b/d+c/a+d/b$

If the numbers were positive, then AM-GM would have killed it without the need for the second condition :rolleyes:

Anyway my solution is tedious, but it works. Let me know if you have a problem somewhere.

We repeatedly use $ac = bd$.

let us first put $\frac{a}{b} = x = \frac{d}{c}$ -----------(1)
and then $\frac{b}{c} = y = \frac{a}{d}$ ----------------(2)

Data says $x + \frac1{x} + y + \frac1{y} = 4$

We want to know the maximum value of
$W = \frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$

But we can use (1) and (2) above to reduce W to
$W = (x + \frac1{x})(y + \frac1{y})$

Now set $x + \frac1{x} = X$ and $y + \frac1{y} = Y$

Our simplified conditions now read.... $
\boxed{X+Y = 4, max(XY) = ??}$

This is simple.... :P
$XY = X(4-X) = 4 - (X^2 - 4X + 4) = 4 - (X-2)^2$
The maximum value is 4
(and it occurs for X=Y=2, further you can substitute it above and find the values for which this happens.However the question does not demand the numbers but only the maximum)

I have left a little bit for you, in particular getting $W = (x + \frac1{x})(y + \frac1{y})$. Try it! :D
• Jan 8th 2008, 02:08 AM
distinct a,b,c,d?
I am a little bit worried about that word distinct because I got a=b=c=d for W = 4.

I think that the maximum for distinct solutions will be when a,b,c and d are infinitely close and that the maximum for W will then be infinitely close to 4. So maybe we don't need to worry.

Very nice proof by the way Isomorphism.
• Jan 8th 2008, 02:50 AM
Isomorphism
Quote:

I am a little bit worried about that word distinct because I got a=b=c=d for W = 4.

I think that the maximum for distinct solutions will be when a,b,c and d are infinitely close and that the maximum for W will then be infinitely close to 4. So maybe we don't need to worry.

Not necessarily. In other words, you got one set of solutions which you have to discard because of the data in the problem.
However note that the problem does not say they are positive real numbers.

So to satisfy
$x + \frac1{x} + y + \frac1{y} = 4$

Solve
$x + \frac1{x} = -3$ and $y + \frac1{y} = 7$
And get some distinct solutions :D

Quote:

Very nice proof by the way Isomorphism.
Thank you (Blush)
• Jan 8th 2008, 03:30 PM
I don't think I explained myself clearly enough the first time so I will have another go.

Quote:

The maximum value is 4
(and it occurs for X=Y=2,
X=2
$x+\frac {1}{x} = 2$
$x^2+1 = 2x$
$x^2-2x+1=0$
x=1
$\frac {a}{b} = 1$
a=b

So the point at which the maximum (4) occurs has a = b, so the numbers are not distinct. This is what I am worried about.

Quote:

x + \frac1{x} = -3 and y + \frac1{y} = 7
If we do this then $XY \not = 4$
• Jan 9th 2008, 04:21 AM
Isomorphism
Quote:

I don't think I explained myself clearly enough the first time so I will have another go.
X=2
$x+\frac {1}{x} = 2$
$x^2+1 = 2x$
$x^2-2x+1=0$
x=1
$\frac {a}{b} = 1$
a=b

So the point at which the maximum (4) occurs has a = b, so the numbers are not distinct. This is what I am worried about.

Oh, yes I understand now :o
You are saying that we have proved the maximum value of W=4 is attained iff a becomes arbitrarily close to b, since they cannot equal. Thus $W \to 4$

Yes, you are right.I missed that point entirely, sorry (Doh)