$a > b > k > 0$ what is bigger: $\frac{a}{b}$ or $\frac{a+k}{b+k}$

Question

The right answer is: : $\frac{a}{b}$ is bigger. But.. I don't understand the intuition behind. Why is it the case? How can I prove it?

Thanks

Answer 1

Increasing the numerator and the denominator by the same amount takes you closer to $1$ (at least as long as all numbers involved are positive). That's the intuition.

Answer 2

$$ak>bk$$ $$ab+ak>ab+bk$$ $$a(b+k)>b(a+k)$$

Can you complete the rest?

Alternative view:

Suppose $a=lb$ where $l>1$ $$a+k=lb+k=l(b+k)-lk$$

$$\frac{a+k}{b+k}=l\left(1-\frac{k}{b+k}\right)=l\left(\frac{b}{b+k}\right)<l$$

Answer 3

$a > b > k > 0$. Note that $\left(~{1 \over a} < {1 \over b} \implies {k \over a} < {k \over b} \implies 1 + {k \over a} < 1 + {k \over b} \implies {1 + k/a \over 1 + k/b} < 1~\right)$:

$$ \color{#f00}{a + k \over b + k} = {1 + k/a \over 1 + k/b}\,{a \over b} \color{#f00}{< {a \over b}} $$