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Proof

(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.
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References

Bibliography

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