Think of mediant fractions

For non-negative integers $n$ define $a(n) = \dfrac{7 + n}{13 + n}$ . Then $a(n) = \dfrac{13 + n - 6}{13 + n} = 1 - \dfrac{6}{13 + n}$ , and so

$a(n + 1) - a(n) = \left(1 - \dfrac{6}{13 + (n + 1)}\right) - \left(1 - \dfrac{6}{13 + n}\right) = \dfrac{6}{13 + n} - \dfrac{6}{14 + n} \gt 0$ .

The proposed inequality is therefore true.

More generally, If $a, b, c$ are positive numbers, then $\frac{ a}{ b} < 1$ if and only if $\frac{ a}{b} < \frac{ a+c}{ b+c}$

Proof: $\frac{ a}{b} < \frac{ a+c}{b+c} \Leftrightarrow a(b+c) < b(a+c) \Leftrightarrow ac < bc \Leftrightarrow \frac{a}{b} < 1$ . $_\square$

Hence, we see that $\frac{ (7+2) } { (13+2) } < \frac{ (7+2) + 1 } { (13 + 2 ) + 1 } = \frac{ 7 + 3 } { 13 + 3 }$ .

Let me illustrate this by using a classroom example. Let's say at the beginning of every geometry class, the teacher gives you a one point "pop quiz". After 13 days, you've only got 7 right. That's when you decide to spend an additional 30 minutes every night to improve these scores. The next day you get it right, so you are at 8/14. The next one is successful too. And again. Your scores are only improving. The inequality remains.

We want to compare which is greater: $\dfrac{7 + 2}{13 + 2}$ or $\dfrac{7 + 3}{13 + 3}$ .

We can evaluate the sums and simplify the fractions to get: $\dfrac{7 + 2}{13 + 2} = \dfrac{9}{15} = \dfrac35$ , and $\dfrac{7 + 3}{13 + 3} = \dfrac{10}{16} = \dfrac{5}{8}$ .

If we make the denominators of both fractions same, then we can compare the numerators to compare the fractions. The LCM of 5 and 8 is 40.

$\dfrac{3}{5} = \dfrac{3 \times 8}{5 \times 8} = \dfrac{24}{40}$ , $\quad \dfrac{5}{8} = \dfrac{5 \times 5}{8 \times 5} = \dfrac{25}{40}$ .

Since $\dfrac{24}{40} < \dfrac{25}{40}$ , we have $\dfrac{3}{5} < \dfrac{5}{8}$ , and the given inequality is indeed true.

generally speaking

suppose a, b, k are positive integers

compare $\frac{a}{b}$ and $\frac{a+k}{b+k}$ by considering the quotion Q = $\frac{\frac{a}{b}}{\frac{a+k}{b+k}}$ = $\frac{a(b+k)}{(a+k)b}$ = $\frac{ab+ak}{ab+bk}$

a > b $\implies$ Q = $\frac{ab+ak}{ab+bk}$ > 1 $\implies$ $\frac{a}{b}$ > $\frac{a+k}{b+k}$

a < b $\implies$ Q = $\frac{ab+ak}{ab+bk}$ < 1 $\implies$ $\frac{a}{b}$ < $\frac{a+k}{b+k}$

therefore the inequality is true

Considering the function $\frac{7+x}{13+x}$ , we notice that it is monotonically increasing in $(-13; +\infty)$ . Therefore the inequality is always true for $x>1$ .