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 $ad+bc \le bd-1.$
 A: 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:)
A: From $\frac ab+\frac cd \lt 1$ and $a+c<n$ it is clear that $a<b;\space c<d$, $n\ge 3$ and $2\le a+c\le n-1$. Furthermore at least one of the two fractions should be less than $\frac 12$ therefore we can put $\frac {a}{2a+k}+\frac cd\lt 1$ where $k\ge 1$ so $$\frac ab+\frac cd\le \frac {a}{2a+1}+\frac cd\lt 1\iff \frac cd\lt \frac{a+1}{2a+1}\qquad (*)$$
Now, fixing $a$, we have for all the possible values of $\frac cd$
$$\frac cd\in \left\{\frac{1}{1+k}\right\}_{k>0}\cup \left\{\frac{2}{2+k}\right\}_{k>0}\cup…..\cup \left\{\frac{n-1-a}{n-1-2+k}\right\}_{k>0}$$ 
and the maximum of $\frac cd$ is in the set 
$$\left\{\frac 12\right\}\cup \left\{\frac 23\right\}\cup…..\cup \left\{\frac{n-1-a}{n-a}\right\}\qquad (**)$$
Without restrictions this maximum is clearly $\left\{\frac{n-1-a}{n-a}\right\}$ but we have to take into account the restriction $(*)$,
Consider the functions $f(x)=\frac{x+1}{2x+1}$ and $g(x)=\frac{x}{x+1}$; $f$ is decreasing and $g$ increasing in its domains of definition. Furthermore the positive solution $x_0$ of the equation $f(x)=g(x)$ gives exactly the value in which $\frac cd=\frac{a+1}{2a+1}$. This solution (unexpectedly it is the Golden ratio) is $x_0 \approx 1.6180$ and to the right of $x_0$ one has $g(x)>f(x)$ (so for example 
$\frac 23\gt \frac {3}{5}$)  

Hence the value of $\frac cd$ we must take is $\frac 12$.
Hence $$\frac ab+\frac cd +\frac {1}{n^3}\le \frac {a}{2a+1}+\frac 12 +\frac{1}{n^3}\le \frac{n-2}{2n-3}+\frac 12+\frac{1}{n^3}=\frac{4n^4-7n^3+4n-6}{4n^4-6n^3}$$
Now, by the absurd, suppose that $$\frac{4n^4-7n^3+4n-6}{4n^4-6n^3}\ge 1\iff 4n\ge n^3+6$$ This is not true for $n\ge 3$.
Thus $$\frac ab+\frac cd +\frac {1}{n^3}\lt 1$$
A: Since $n>a+c \ge 2$ then $n \ge 3$. Moreover, $a<b$ and $c<d$, because $$\frac ab + \frac cd <1$$.
Consider the following cases.
a) Let $b \ge n$ and $d \ge n$, then $$\frac ab + \frac cd \le \frac an + \frac cn=\frac {a+c}{n} \le \frac {n-1}{n}=1-\frac 1n<1- \frac{1}{n^3}.$$
b) Let $b \le n$ and $d \le n$, then 
$$\frac ab + \frac cd <1 \Rightarrow ad+bc<bd \Rightarrow ad+bc+1 \le bd$$
From here
$$\frac ab + \frac cd \le 1- \frac{1}{bd} \le 1- \frac {1}{n^2}<1- \frac{1}{n^3}$$
c) Let $b < n<d$. If $d \le n^2$. then $bd \le n^3$ and then 
$$\frac ab + \frac cd \le 1- \frac{1}{bd} <1- \frac{1}{n^3}$$
If $d>n^2$, then 
$$\frac cd \le \frac{n-2}{n^2}=\frac 1n- \frac {2}{n^2},$$
because $c<n-a \le n-1$, that $c \le n-2$. Suppose 
$$\frac ab + \frac cd \ge 1 - \frac {1}{n^3}.$$
Then 
$$1-\frac ab \le \frac cd + \frac {1}{n^3} \le \frac 1n - \frac {2}{n^2}+\frac{1}{n^3}< \frac 1n. $$
Hence, what $b>n(b-a)\ge n$ (here we take into account that $a<b$), which contradicts inequality $b<n<d$.
d) Let $d < n<b$. If $b \le n^2$, then $bd < n^3$, and then 
$$\frac ab +\frac cd \le 1 - \frac {1}{bd} < 1- \frac {1}{n^3}$$
If $b>n^2$, then
$$\frac ab \le \frac {n-2}{n^2}=\frac 1n - \frac {2}{n^2},$$
because $a<n-c \le n-1$, that $a \le n-2$. Suppose 
$$\frac ab + \frac cd \ge 1 - \frac {1}{n^3}.$$
Then 
$$1 - \frac cd \le \frac ab + \frac {1}{n^3} \le \frac 1n - \frac {2}{n^2}+\frac{1}{n^3}< \frac 1n.$$
$c<d \Rightarrow d> n(d-c) \ge n$. Got a contradiction with inequality $d<n<b$.
So inequality $$\frac ab + \frac cd < 1 - \frac {1}{n^3}$$ is done in all cases that had.
prove.
