Which is the larger,

$$
\frac{1998^{2000}}{2000^{1998}} \quad \text { or } \quad \frac{1997^{1999}}{1999^{1997}} ?
$$
megoI. solution. We will show that if $n>0$ is an integer, then

$$
\frac{(n+1)^{n+3}}{(n+3)^{n+1}}>\frac{n^{n+2}}{(n+2)^{n}}
$$

This generalization implies that the solution to our problem (for $n=1997$) is:

$$
\frac{1998^{2000}}{2000^{1998}}>\frac{1997^{1999}}{1999^{1997}}
$$

Multiply both sides of (1) by $(n+3)^{n+1}(n+2)^{n}$: since we are dealing with positive integers, the direction of the inequality does not change:

$$
(n+1)^{n+3}(n+2)^{n}>n^{n+2}(n+3)^{n+1}
$$

Rewriting this identically:

$$
(n+1)^{n}(n+2)^{n}(n+1)^{3}>n^{n}(n+3)^{n} \cdot n^{2}(n+3),[(n+1)(n+2)]^{n}(n+1)^{3}>[n(n+3)]^{n} \cdot n^{2}(n+3)
$$

This holds due to the following:

$$
(n+1)(n+2)=n^{2}+3 n+2>n^{2}+3 n=n(n+3), \quad \text{so} \quad [(n+1)(n+2)]^{n}>[n(n+3)]^{n}
$$

since $n>0$ is an integer. Therefore,

$[(n+1)(n+2)]^{n}(n+1)^{3}>[n(n+3)]^{n}(n+1)^{3}=[n(n+3)]^{n}\left(n^{3}+3 n^{2}+3 n+1\right)>>[n(n+3)]^{n}\left(n^{3}+3 n^{2}\right)=[n(n+3)]^{n} n^{2}(n+3)$

The direction of the relation remained the same due to the equivalent transformations, thus proving (1).

Somogyi Dávid (Fazekas M. Főv. Gyak. Gimn., 9. grade)

II. solution. We will prove the following lemma: Let $a$ and $b$ be positive integers, then

$$
\left(\frac{a+1}{b+1}\right)^{b+1}>\left(\frac{a}{b}\right)^{b}
$$

Indeed, write the inequality between the arithmetic and geometric means for the numbers $1, \underbrace{\frac{a}{b}, \frac{a}{b}, \ldots, \frac{a}{b}}_{b \text{ times}}$:

$$
\frac{1+\frac{a}{b}+\frac{a}{b}+\cdots+\frac{a}{b}}{b+1} \geq \sqrt[b+1]{1 \cdot \frac{a}{b} \cdot \frac{a}{b} \cdots \cdot \frac{a}{b}}
$$

$\underbrace{\text { i.e. }}_{b \text{ times }}$

$$
\frac{1+a}{b+1} \geq \sqrt[b+1]{\left(\frac{a}{b}\right)^{b}}, \quad \text { and thus indeed } \quad\left(\frac{a+1}{b+1}\right)^{b+1} \geq\left(\frac{a}{b}\right)^{b}
$$

and equality holds only if $1=\frac{a}{b}$, i.e., if $a=b$. For $a=1997, b=1999$, we have

$$
\left(\frac{1998}{2000}\right)^{2000}>\left(\frac{1997}{1999}\right)^{1999}
$$

Multiply both sides of the inequality by $2000^{2}$:

$$
\frac{1998^{2000}}{2000^{1998}}=\left(\frac{1998}{2000}\right)^{2000} \cdot 2000^{2}>\left(\frac{1997}{1999}\right)^{1999} \cdot 2000^{2}>>\left(\frac{1997}{1999}\right)^{1999} \cdot 1999^{2}=\frac{1997^{1999}}{1999^{1997}}
$$

thus proving that

$$
\frac{1998^{2000}}{2000^{1998}}>\frac{1997^{1999}}{1999^{1997}}
$$

Hegyi Márta (Budapest, Szent István Gimn., 10. grade)

Remarks. 1. Several students used calculators, but the machine can only display a certain number of decimal places, and calculations with rounded values can reverse the relation - this happened in some cases! Kiss Norbert

(Budapest, Szent István Gimn., 10. grade) also examined the effect of rounding errors on the result, making his solution completely correct.

2. Similarly, Andrássy Zoltán (Fazekas M. Főv. Gyak. Gimn., 10. grade) provided a good solution by examining the effect of rounding errors, using the base-10 logarithms of the fractions.
3. A typical mistake in some of the solutions was to conclude from

$$
\frac{1998}{2000}>\frac{1997}{1999} \text { that } \quad\left(\frac{1998}{2000}\right)^{1998}>\left(\frac{1997}{1999}\right)^{1997}
$$

which is not justified, since the powers of positive numbers less than 1 decrease with increasing exponents, e.g.,

$$
\left(\frac{1997}{1999}\right)^{1998}<\left(\frac{1997}{1999}\right)^{1997}
$$