<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] 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 $$\frac ab+\frac cd +\frac {1}{n^3}\lt 1$$
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