Assuming that a b </mfrac> &lt; c d </mfrac> &lt; 1 , a

Ryan Reynolds

Ryan Reynolds

Answered question

2022-05-20

Assuming that a b < c d < 1, arrange b a , d c , b d a c , b + d a + c , and 1
My attempt
I've been able to get three of them in position ,namely
b d a c > b a > d c > 1
since it's given that a b < c d from which I have that b a > d c
and since b a , d c > 1 I have that b d a c > b a > d c > 1
Now I am kind of clueless on how to tackle the expression b + d a + c ,as I don't see how I can get this by manipulating any of the given relations.

Answer & Explanation

nifeonibonitozg

nifeonibonitozg

Beginner2022-05-21Added 12 answers

From
b = a b a > a d c
it follows that
b + d a + c > a d c + d a + c = d c
and in a similar way you get
b + d a + c < b a .
There is also a general theorem which states that for positive numbers,
min a i b i a 1 + . . . + a n b 1 + . . . + b n max a i b i
and this is proved using the same ideas.
Alani Conner

Alani Conner

Beginner2022-05-22Added 3 answers

Just to continue from where you stopped b + d a + c is between d c and b a . I guessed it using an example, and then proved the general result.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?