prove that if b,d>0 and a/b<c/d, then a/b<(a+c)/(b+d)<c/d?

Answer 1

Assume b, d > 0, and assume (a/b) < (c/d).

{We want to prove that (a/b) < (a + c)/(b + d) < (c/d). }

Since (a/b) < (c/d), if we multiply both sides by bd (Note: this shouldn't change the inequality sign, since b > 0 and d > 0), we obtain

ad < bc

Adding ab to both sides,

ab + ad < ab + bc

Factoring both sides,

a(b + d) < b(a + c)

Dividing both sides by b(b + d) gives us

a/b < (a + c)/(b + d) {Part I of proof done}

As already stated,

ad < bc

Adding cd to both sides, we obtain

ad + cd < bc + cd

Factoring both sides,

d(a + c) < c(b + d)

Dividing both sides by d(b + d), we get

(a + c)/(b + d) < c/d {Part II of proof done}

Combining parts I and II, since

(a/b) < (a + c)/(b + d)

AND

(a + c)/(b + d) < c/d

then these two inequalities mean

(a/b) < (a + c)/(b + d) < (c/d)

Answer 2

Note that you can write:

(a+c)/(b+d) = [(a/b)(1/d) + (c/d)(1/b)]/[(1/d) + (1/b)], upon dividing top and bottom by bd.

So the fraction in the middle is just a weighted average of (a/b) and (c/d), and since the weights (1/d and 1/b) are positive, this average must lie strictly between a/b and c/d.

Answer 3

Proof by contradiction - suppose
a/b >= (a+c)/(b+d) - denominators are positive, so cross-multiply:
→ a(b+d) >= b(a+c)
→ ab+ad >= ab+bc
→ ad >= bc
→ a/b >= c/d. Contradiction.

Similarly for the right-hand inequality.