Problem 1198

Instead of (2), we need to see the appropriate inequality between the values reduced by 1 on both sides:

$$\frac{d-**}{**+n-2}>-\frac{b}{**+n-1}$$

This holds if and only if, by subtracting the right side from the left, we get a positive value. The denominators are positive because $** \geqq 0$ and $a>2$, so it suffices to determine the sign of the numerator of the difference. The numerator is:

$$(d-**)(**+n-1)+b \cdot(**+n-2)=(d-**+b)(**+n-1)-b$$

The expression in the first parentheses can be transformed using (1) as follows:

$$d-**+b=\left(\frac{**+1-b}{**+1}-1\right) \cdot**+b=b\left(1-\frac{**}{**+1}\right)=\frac{b}{**+1}$$

Thus, the expression to be examined can be written in the following form:

$$b\left(\frac{**+n-1}{**+1}-1\right)=\frac{b(n-2)}{**+1}$$

which is indeed positive because $b$ and $**+1$ are positive and $n>2$.

We did not need to use that $n$ is a natural number, only that it is greater than 2, and we did not need the relationship between $b$ and $**$, only that $** \geqq 0$.

Péter Takács (Budapest, Berzsenyi D. Gymnasium II. o. t.)