## 22. Math Puzzle $3 / 67$

Peter wants to use a balance scale, whose beam lengths $a$ and $b$ are no longer exactly equal. He first places a $5 \mathrm{~kg}$-"weight" on the left pan and weighs, and then the $5 \mathrm{~kg}$ piece on the right pan and weighs the rest of the apples.

Is the weighed amount heavier or lighter than $10 \mathrm{~kg}$?
With the lever law, for the first weighing: $5: F_{2}=a: b \rightarrow F_{2}=5 \frac{a}{b}$ and for the second weighing $F_{1}=5 \frac{a}{b}$. Thus, the weight of the entire amount is

$$
G=F_{1}+F_{2}=5 \cdot \frac{a^{2}+b^{2}}{a b}
$$

Since $a^{2}+b^{2}>2 a b$ (because $a^{2}-2 a b+b^{2}=(a-b)^{2}>0$), the numerator is greater than the denominator. Therefore, it is proven that the weighed amount is greater than $10 \mathrm{~kg}$.