Prove that a b < a + c b + d < c d I'm...

Erin Lozano

Erin Lozano

Answered

2022-06-25

Prove that a b < a + c b + d < c d
I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let a , b , c , d be positive real numbers, prove that if a b < c d , then a b < a + c b + d < c d

Answer & Explanation

kpgt1z

kpgt1z

Expert

2022-06-26Added 23 answers

HINT: Note that
a + c b + d a b = a + c b + d b b a b b + d b + d = b ( a + c ) a ( b + d ) b ( b + d ) = a b + b c a b a d b ( b + d ) = b c a d b ( b + d ) .
You’d like to show that this difference is positive, so you’d like to know that b c > a d. Use the fact that a b < c d to show this.
The other inequality can be handled similarly.
arridsd9

arridsd9

Expert

2022-06-27Added 12 answers

a b = a ( 1 + d b ) b ( 1 + d b ) = a + d ( a b ) b + d < a + d ( c d ) b + d = a + c b + d
and similarly for the other inequality.

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