<s>[INST] 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 $ad+bc \le bd-1.$
 [/INST] 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.
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