[ [trigonometric inequalities ]

Solve the inequality $\sin \frac{x}{x^{2}+1}+\frac{x^{2}+1}{x} \cos \frac{x}{x^{2}+1}>0$
1) Note that $x \neq 0$ and $t=\frac{x}{x^{2}+1}=\frac{1}{x+\frac{1}{x}}$. Since $|x+1 / x| \geq 2$, then $|t| \leq 1 / 20$, so $00$ and $\cos t>0$. Moreover, $1 / t>0$, which means $\sin t+1 / t \cos t>0$. Therefore, for all $x>0$, the original inequality holds.
3) If $x>0$, then $-1 / 2 \leq t0.1 / t [ [various and GMT on the coordinate plane $]$  [ Properties and characteristics of a parallelogram ]  Author: Kalinin A. At the same time, Anya and Borya (their speeds are constant but not necessarily the same) walked out from villages A and B towards each other. If Anya had left 30 minutes earlier, they would have met 2 km closer to village B. If Borya had left 30 minutes earlier, the meeting would have taken place closer to village A. By how much?

## Solution

Suppose Kolya (30 minutes later than Borya) and Tolya (30 minutes earlier than Borya) also left from B with the same speed, and Tanya (30 minutes earlier than Anya) left from A.

First method. Anya meets Tolya, Borya, and Kolya sequentially at equal time intervals. Therefore, the distances between the meeting points of Anya with the three boys are equal. It is clear that the meeting of Anya with Kolya happened 30 minutes later, and at the same point where the meeting of Tanya with Borya mentioned in the condition took place. Thus, the distances between the meeting points of Anya with the three boys are 2 km, and, in particular, the meeting point of Anya with Tolya is 2 km closer to A than the meeting point of Anya with Borya.

Second method. Consider the graphs of the movements of Anya, Borya, Tanya, and Tolya. These are two pairs of parallel lines, so the points of their intersections are the vertices of a parallelogram. Two opposite vertices have the same ordinate (the meeting point of Anya with Borya coincides with the meeting point of Tanya with Tolya), which means that this point is the arithmetic mean of the ordinates of the other two vertices. Therefore, the other two meeting points lie on opposite sides but at the same distance of 2 km from this point.

## Answer

By 2 km.