Proof of inequality in $\mathbb{R^+}$

Question

I have this statement:

You have this fraction: $\frac{p}{q}$, where:

$q > p$ and $q,p \in \mathbb{R^+}$

Proof if $\frac{p}{q} < \frac{p+1}{q+2}$

Basically my development was:

$i)$ $\frac{p}{q}$ has an interval of $(0, 1)$

$i.a) \frac{p+1}{q+2} = \frac{p}{q+2} + \frac{1}{q+2}$

$ii)$ We can notice that $\frac{p}{q} > \frac{p}{q+2}$

$iii)$ According with the last. Let's suppose that $\frac{p}{q} < \frac{p+1}{q+2}$ is true. That is true, if and only if, $\frac{1}{q+2} < \frac{p}{q} - \frac{p}{q+2}$.

I think the step $ iii) $ of this way:

Let $k = \frac{p}{q} - \frac{p}{q+2}$, therefore $\frac{p}{q}$ is $k$ units bigger than $\frac{p}{q+2}$, to "counteract" this, i need that $\frac{1}{q+2}$ be greater than the difference $ k $(from step $i.a)$, and therefore be bigger than $\frac {p} {q}$

$iv)$ Assuming that $\frac{1}{q+2} < \frac{p}{q} - \frac{p}{q+2}$, i will multiply it by $(q+2)$, now i have:

$1 + p< \frac{p(q+2)}{q}$, that is:

$1 + p< \frac{p}{q} *^(q+2)$

I'll write the interval of the left and right side of this inequality.

$[1, \infty^+) < (0, 1) * [2, \infty^+)$ ( assuming that $p,q$ they are infinitely small positives)

Until here I have arrived. In fact, I do not know if it is correct (if it is NOT, I would like to know why), otherwise I would like to know how to continue.

Thanks in advance.

Answer 1

We have that

$$\frac{p}{q} < \frac{p+1}{q+2}\iff p(q+2)<q(p+1) \iff pq+2p<pq+q \iff 2p<q$$

Answer 2

We have $$\frac{p}{q} < \frac{p+1}{q+2}\quad\iff\quad p(q+2) = pq + 2p< q(p+1) = pq + q\quad\iff\quad 2p < q\ ,$$ where we used the fact that $q>0$.

Answer 3

You cannot prove it, since it is false. Take $p=2$ and $q=3$. It is true that $p<q$, but it is false that $\frac pq<\frac{p+1}{q+2}$.

Answer 4

Let $2p\ge q$. For instance, $p=2, q=3$. Then $\frac{p+1}{q+2}=\frac35\lt\frac23$.