Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$  positive integers $\Rightarrow a+c \le n-1$
 and $ad+bc \le bd-1.$

For convenience, let us to think that $a\le c$ and $t=d-c\in\mathbb{N}$.
1. Not hard to see that if $\frac{a}{b}+\frac{c}{c+t}<1$ then
\begin{equation}
\frac{a}{b}+\frac{c}{c+t}\le\frac{a}{\left\lfloor  \frac{ac}{t}\right\rfloor+a+1} + \frac{c}{c+t} < 1.
\end{equation}
2. Now, let us fix only $a\le c$ and let $t\ge1$ is variable. We can show that  the sum
$$\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}$$
has the maximal value for $t=1$, or by other words
$$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1}=\frac{1}{(c+1)(a(c+1)+1)}.$$
2.a Let $t\le c^2$. Note, that $\left\lfloor \frac{ac}{t}\right\rfloor =  \frac{ac}{t} - \frac{ac(\mathrm{mod\ }t)}{t}\ge   \frac{ac}{t} - \frac{t-1}{t}$, so
$$\frac{t}{c+t}-\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1}\ge\frac{t}{c+t}-\frac{at}{a(t+c)+1} = \frac{t}{(c+t)(a(t+c)+1)},$$
and we have to prove that
$$\frac{t}{(c+t)(a(t+c)+1)}\ge\frac{1}{(c+1)(a(c+1)+1)}$$
or
$$(t-1)\left(t-\frac{ac^2+c}{a}\right)\le0$$
or
$$t < c^2+\frac{c}{a}.$$
Case 2.a is done.
2.b Now, let $t \ge c^2+1$. Since $a\le c$, $\left\lfloor\frac{ac}{t}\right\rfloor=0$, therefore we have to prove that
$$\frac{t}{c+t}-\frac{a}{a+1}\geq\frac{1}{c+1}-\frac{a}{a(c+1)+1 }.$$
In view of $t\ge c^2+1$, it suffices to prove
$$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}.$$
It is easy to see that for the function $f(x)=\frac1x$ the following property is true:


*

*If $y\ge x>0$ then for all $\delta > 0$ : $f(x)-f(x+\delta)\ge f(y)-f(y+\delta)$.
From the last statement and from $\frac{1}{c}+c-a\geq \frac{1}{a}$ we can conclude
$$\frac{1}{1+a}-\frac{1}{1+c}\ge\frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{c}+(c-a)}\ge \frac{1}{1+c+\frac{1}{c}}-\frac{1}{1+c+\frac{1}{a}}$$
Now 2. is done.
3. From 1. and 2. :
$$\frac{a}{b}+\frac{c}{c+t}\le \frac{a}{\left\lfloor\frac{ac}{t}\right\rfloor+a+1}+\frac{c}{c+t}\le \frac{a}{a(c+1)+1}+\frac{c}{c+1}=1-\frac{1}{(c+1)(a(c+1)+1)}\le 1-\frac{1}{(a+c)^3} < 1-\frac{1}{n^3}$$
I hope that more laconic solvation exists:)
