(related to Theorem: Inequality of the Arithmetic Mean)

- By hypothesis, the two real numbers $x,y$ are ordered as follows $x < y.$
- Because $2x=x+ x$ and because of 3rd rule of calculations with inequalities we have $2x < x + y.$
- For the same argument $x + y < 2y.$
- Multiplying both inequations by $\frac 12$ (6-th rule of calculations with inequalities) give us $x < \frac{x+y}2 < y$ for the arithmetic mean.∎

**Heuser Harro**: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition