0

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

YohanRoth
  • 1,437
  • Think intuitively as $k \to \infty$. The fraction goes to $1$, clearly. But $\frac ab > 1$ so $f(k) = \frac{a + k}{b + k}$ is decreasing. – MathematicsStudent1122 Oct 23 '16 at 20:36
  • As an alternative argument, note that cross multiplying shows that the desired inequality is equivalent to $a(b+k)>b(a+k)$ or $ak>bk$. – lulu Oct 23 '16 at 20:38

3 Answers3

3

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.

Arthur
  • 199,419
2

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

Can you complete the rest?

Alternative view:

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

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

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
0

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

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

Felix Marin
  • 89,464