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We can prove that average of two numbers $a,b$ where $a<b$ will be between $a$ and $b$ as follows

$a < b$
$a + a < a + b$
a < $\dfrac{a + b}{2 }$

$a < b$
$a + b < b + b$
$\dfrac{a + b}{2 }< b$

Thus
$a < \dfrac{a + b}{2 }< b$

Similarly, is it true always and can be proved, that average of three numbers, $a,b,c$ where $a<b<c$ will be between $a$ and $b$ if $b-a$ > $c-b$

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Yes, it is always true.

From what you've already got, we have $$a<b<\frac{b+c}{2}<c\implies a<\frac{b+c}{2}$$ $$\implies 2a<b+c\implies 3a<a+b+c\implies a<\frac{a+b+c}{3}$$

Also, we have $$b-a>c-b\implies a+c\lt 2b\implies a+b+c\lt 3b\implies \frac{a+b+c}{3}\lt b$$

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