Can You Prove this Specific Inequality Involving Positive Integers?

[anemone]

Positive integers $l,\,k,\,m,\,n$ satisfying $l+m \le 1982$ and $\dfrac{l}{k}+\dfrac{m}{n}<1$. Prove that $1-\dfrac{l}{k}-\dfrac{m}{n}>\dfrac{1}{1983^3}$.

 

[anemone]

Hint:

Spoiler

Let $1-\dfrac{l}{k}-\dfrac{m}{n}=\dfrac{a}{kn}$ for some positive integer $a$. And see what happens if we let $a>n$ and $a\le n$.

The skill of using the concept of contradicting of minimality with respect to certain variable helps too.

 

[anemone]

Solution of other:

Spoiler

Let us consider the general case with 1982 replaced by $P$. We have $l
\dfrac{1}{P}$. So there is no need in making $n$ too big.

Certainly we must have $n<2Pm<2P^2$. Similarly for $k$. So there are only finitely many candidates for $l,\,m,\,n,\,k$. Hence there is a set of values which minimizes $1-\dfrac{l}{k}-\dfrac{m}{n}$. Let us adopt these values.

Clearly $1-\dfrac{l}{k}-\dfrac{m}{n}=\dfrac{a}{kn}$ for some positive integer $a$. We may assume that $k>n$. The fact that $l,\,m,\,n,\,k$ is an optimal set means that $a$ cannot be too large. Multiplying across, $a=kn-ln-km$. So if $a>n$, we could increase $l$ by 1. That would reduce $1-\dfrac{l}{k}-\dfrac{m}{n}$ to $\dfrac{a-n}{kn}$, contradicting minimality. So, $a\le n$.

Now, $kn=a+ln+km\le a+lk+km=a+k(l+m)\le a+kP$. Hence, $n\le P+\dfrac{a}{k}\le P+1$.

But $\dfrac{a}{kn}+\dfrac{l}{k}=1-\dfrac{m}{n}\ge \dfrac{1}{n}\ge \dfrac{1}{P+1}$.

We have $\dfrac{a}{kn}+\dfrac{l}{k}=\dfrac{1}{k(l+1)}\le \dfrac{1}{k(l+1)}\le \dfrac{1}{kP}$. So, $\dfrac{1}{k}\ge \dfrac{1}{P(P+1)}>\dfrac{1}{(P+1)^2}$.

Hence $1-\dfrac{l}{k}-\dfrac{m}{n}=\dfrac{a}{kn}>\dfrac{a}{(P+1)^3}\ge \dfrac{1}{(P+1)^3}=\dfrac{1}{1983^3}$.

 

[anemone]

I am thinking it would be nice to show another variant of the same problem that uses other method to tackle it...here goes the problem:

Spoiler

Let $l,\,k,\,m,\,n\in\Bbb{Z_+}$ and $a=1-\dfrac{l}{k}-\dfrac{m}{n}$. If $a>0$ and $l+m\le 1982$, then prove that $a>\dfrac{1}{1983^3}$.

We have 3 cases to analyze:

Case I: If $k,\,n\ge 1983$, then $a\ge 1-\dfrac{l+m}{1983}\ge 1-\dfrac{1982}{1983}>\dfrac{1}{1983^3}$.

Case II: if $k,\,n\le 1983$, then $a=\dfrac{kn-ln-km}{kn}>0$ so that $kn-ln-km \ge 1$. Thus, $a\ge \dfrac{1}{kn}>\dfrac{1}{1983^3}$.

Case III:

Suppose now that $k<1983<n$, if $n>1983^2$ and $a<\dfrac{1}{1983^3}$, then $\dfrac{m}{n}<\dfrac{1982}{1983}$, thus, $1-\dfrac{l}{k}<\dfrac{1}{1983}$. This implies that $k>1983(k-l)>1983$ because $l\le k+1$, which is absurd. If $n\le 1983^2$, then as in the case II above, we have $a\ge \dfrac{1}{kn}>\dfrac{1}{1983^3}$.

 

[CellCoree]

To prove the given statement, we will use the given conditions and properties of inequalities. First, we will rearrange the given inequality to get:

$1-\dfrac{l}{k}-\dfrac{m}{n}>\dfrac{1}{1983^3}$

We can then multiply both sides by $k$ and $n$ to get:

$k\cdot n - l\cdot n - k\cdot m > \dfrac{k\cdot n}{1983^3}$

Next, we can use the given condition that $l+m \le 1982$ to substitute $l$ with $1982-m$ in the above inequality. This gives us:

$k\cdot n - (1982-m)\cdot n - k\cdot m > \dfrac{k\cdot n}{1983^3}$

Simplifying this, we get:

$1982\cdot n - k\cdot m > \dfrac{k\cdot n}{1983^3}$

Since $l+m \le 1982$, we know that $k\cdot m \le k\cdot (1982-m) = 1982k - k\cdot m$. This means that $1982\cdot n - k\cdot m \ge 1982\cdot n - (1982k - k\cdot m) = 1982\cdot n - 1982k + k\cdot m$. Substituting this into our previous inequality, we get:

$1982\cdot n - 1982k + k\cdot m > \dfrac{k\cdot n}{1983^3}$

Next, we can use the given condition that $\dfrac{l}{k} + \dfrac{m}{n} < 1$ to substitute $\dfrac{l}{k}$ with $1-\dfrac{m}{n}$ in the above inequality. This gives us:

$1982\cdot n - 1982k + k\cdot (1-\dfrac{m}{n}) > \dfrac{k\cdot n}{1983^3}$

Simplifying this, we get:

$1982\cdot n - 1982k + k - k\cdot \dfrac{m}{n} > \dfrac{k\cdot n}{1983^3}$

Finally, we can use the property that $k\cdot