4. (13 points) A triathlon competitor swam 1 km in the first stage. In the second stage, he cycled 25 km, and in the third stage, he ran 4 km. He completed the entire distance in 1 hour 15 minutes. Before the competition, he tried out the course: he swam for $1 / 16$ hours, cycled and ran for $1 / 49$ hours each, covering a total of $5 / 4$ km. During the competition, he completed each stage at the same speed as during the training. How long did he cycle and at what speed?
Answer: $5 / 7$ hours; 35 km/h.

Solution. Let $v_{1}, v_{2}, v_{3}$ be the speeds of the athlete on stages $1, 2, 3$ respectively. From the condition, we have: $\frac{1}{v_{1}}+\frac{25}{v_{2}}+\frac{4}{v_{3}}=\frac{5}{4}$ hours; $\frac{1}{16} v_{1}+\frac{1}{49} v_{2}+$ $\frac{1}{49} v_{3}=\frac{5}{4}$ km. Adding these equations and considering that for any positive numbers $x, y$ the inequality $x+y \geq 2 \sqrt{x y}$ (1) holds, we get:

$$
\frac{5}{2}=\left(\frac{1}{v_{1}}+\frac{1}{16} v_{1}\right)+\left(\frac{25}{v_{2}}+\frac{1}{49} v_{2}\right)+\left(\frac{4}{v_{3}}+\frac{1}{49} v_{3}\right) \geq
$$

$$
\geq 2 \sqrt{1 \cdot \frac{1}{16}}+2 \sqrt{25 \cdot \frac{1}{49}}+2 \sqrt{4 \cdot \frac{1}{49}}=2\left(\frac{1}{4}+\frac{5}{7}+\frac{2}{7}\right)=\frac{5}{2} \text {. }
$$

Equality is achieved if and only if the terms on the left side of inequality (1) are equal. Therefore,

$$
\frac{1}{v_{1}}=\frac{1}{16} v_{1} ; \quad \frac{25}{v_{2}}=\frac{1}{49} v_{2} ; \quad \frac{4}{v_{3}}=\frac{1}{49} v_{3}
$$

which means $v_{1}=4 \frac{\text { km }}{4}, v_{2}=35 \frac{\text { km }}{\text { h }}, v_{3}=14 \frac{\text { km }}{\text { h }}$.

Evaluation criteria. Correct answer with full justification 13 points. There are arithmetic errors with the correct solution approach minus 2 points. Correct system of equations obtained - 2 points (1 point for each). Inequality (1) or an equivalent one used and consequence (2) obtained - another 6 points (total 8 points). Conclusion that equality holds in all three cases in (1) - another 4 points (total 12 points).