Which expressions is greater, $\frac{b}{c+d+b}$ or $\frac{1}{3}$?

Question

There are 3 kinds of animals: Cats, dogs, bears. # of cats is $c$, # of dogs is $d$ and # of bears is $b$. It's said that $$c>d>b.$$ The question is:

Which expression is greater $\dfrac{b}{c+d+b}$ or $\dfrac{1}{3}$?

The thing that I didn't understand is what $\dfrac{b}{c+d+b}$ actually means. And how can I solve for it?

Answer 1

Since $$c\gt d\gt b$$ we have $$c+d+b\gt b+b+b=3b$$ As $c,d,b\ge 0$, $$\frac{1}{c+d+b}\lt\frac{1}{3b}$$ $$\frac{b}{c+d+b}\lt\frac{b}{3b}$$ $$\frac{b}{c+d+b}\lt\frac{1}{3}$$

Answer 2

To properly show the inequality, we would need a bit of algebra. That's covered in the other answer. I want to focus on why you should believe this is true.

If there are $b$ bears, $c$ cats, and $d$ dogs, then the ratio $r = \dfrac{b}{b + c + d} = \dfrac{\text{number of bears}}{\text{number of animals}}$ is just asking for the fraction of animals that are bears.

Since we have the least number of bears, at most $\frac{1}{3}$ of the animals can be bears; $r \le \frac{1}{3}$. But since we have strictly fewer bears than any other animal, we must have $r < \frac{1}{3}$.

Answer 3

I prefer to keep things in fractional form, and to try to get the same numerator to compare.

$$\frac{b}{c+d+b} = \frac{b}{c+d+b}\cdot\frac{\frac{1}{b}}{\frac{1}{b}}= \frac{\frac{b}{b}}{\frac{c+d+b}{b}}=\frac{1}{\frac{c}{b}+\frac{d}{b}+1}\lt\frac{1}{3}$$