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**Soroban** Hello, ice_syncer!

An interesting problem . . .

Let $\displaystyle \frac{a}{b}$ be any positive rational number less than $\displaystyle \sqrt{2}$

We have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2}$ .[1]

. . Add 1 to both sides: .$\displaystyle \frac{a}{b} + 1 \:<\:\sqrt{2} + 1$

. . Take reciprocals: .$\displaystyle \frac{1}{\frac{a}{b} + 1} \:>\:\frac{1}{\sqrt{2}+1} $

. . Add 1 to both sides: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \:>\:1 + \frac{1}{\sqrt{2}+1} $ .[2]

The left side is: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \;=\;1 + \frac{1}{\frac{a+b}{b}} \;=\;1 + \frac{b}{a+b} \;=\;\frac{a + 2b}{a+b}$

The right side is: .$\displaystyle 1 +\frac{1}{\sqrt{2}+1}\!\cdot\!{\color{red}\frac{\s qrt{2}-1}{\sqrt{2}-1}} \;=\;1 + \frac{\sqrt{2} - 1}{1} \;=\;\sqrt{2}$

. . Hence, [2] becomes: .$\displaystyle \frac{a+2b}{a+b} \:> \:\sqrt{2}$ .[3]

Combining [1] and [3], we have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2} \:<\:\frac{a+2b}{a+b} $