If $$a<b<c<d$$ Which is larger without using examples of numbers?

$$x=(a+b)(c+d), y=(a+c)(b+d), z=(a+d)(b+c)$$

I did this exercise, but I have not managed to complete it. The first thing I did was to expand them and then compare ax with z and so on, I know that the correct answer is that z is greater, but I can not assume I have to prove it and my doubt is at the end of taking away the similar terms, since I am assuming that the difference of one side is less than the term that remains for the other.


In order to solve this you can expand the three expressions, and then compute $z-x$ and $z-y$ which are respectively equal to $(d-b)(c-a)$ and $(d-c)(b-a)$ which are both positive. Therefore $z$ is the largest of all three variables.


Here is a proof $x \lt y$

$$ x = ac + ad + bc + bd$$

$$ y = ab + ad + cb + cd$$

Let us start with assuming $x < y$

$$ x = ac + ad + bc + bd \lt ab + ad + cb + cd = y$$

Cancelling out like terms:

$$ ac + bd \lt ab + cd $$

And then rearranging:

$$0 \lt cd - ac + ab -bd$$

$$ 0 \lt c(d-a) +b(a-d)$$

For simplicity, let's define g = d-a


$$0 \lt cg - bg$$

$$ 0 \lt g(c-b)$$

Where $g$ is positive and $c-b$ is positive, so indeed

$$ 0 \lt g(c-b)$$

is true and thus, the assumption $x \lt y$ is true.


Hint: assuming $$x\le y$$ we get by expanding $$ac+bc+ad+bd\le ab+bc+ad+cd$$ then we get $$c(a-d)\le b(a-d)$$ so we get $$(a-d)(c-b)\le 0$$ Can you finish? since $a-d<0$ and $c-b>0$ then our inequality is true.

  • $\begingroup$ In that part I stayed because by solving them how can I know that ab +cd>ac+bd? $\endgroup$ – Sonia f May 3 '18 at 15:53
  • $\begingroup$ And I think there's an error in what you said instead of a-d it would be a-b $\endgroup$ – Sonia f May 3 '18 at 15:54
  • $\begingroup$ Something went wrong. (a-d) is negative and (c-d) is negative. Which means (a-d)(c-d)>0. Which is a contradiction, which implies the original assumption x<y is false. However, testing some real numbers shows x<y is true. $\endgroup$ – CEP May 3 '18 at 16:01
  • $\begingroup$ @sonia $ a-d $ is correct. In the first inequality cancel out the terms. Then rearrange it by subtracting the right terms. $\endgroup$ – Patrick Abraham May 3 '18 at 16:03

Let $c-b = k>0$

Then $y=(a+c)(b+d) = $

$(a+b+k)(c + d -k) = $

$(a+b)(c+d) + k[c+d - a-b] -k^2=$

$x + k[d-a + k] - k^2=$

$x + k(d-a) > x$.

So $y > x$

Let $d-c = m>0$

$z = (a+d)(b+c)=$

$(a + c+m)(b+ d - m)=$

$(a+c)(b+d) + m[b+d - a-c] - m^2=$

$y + m[b-a +m] - m^2 =$

$y + m(b-a) > y$.

So $z > y$

So $z > y > x$.


Also there is AM-GM

If $j < k,m; k,m < n$ then $nj > km$ so

$(a+b) < (a+c)$ and $(b+d)< (c+d)$ so $(a+b)(c+d) > (a+c)(b+d)$... but ... I don't know. That doesn't have as "hands on" conviction. (Although its really the same thing.


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