# Show that $a \lt \frac{a + b}{2} \lt b$ for $a\lt b$ and $a, b \in \mathbb{R}$

How can I prove this statement true?

I have tried saying starting like this:

$a = 0; \qquad b>0.$

But I don't know where to proceed from here

• from x<y you know x/2 < y/2. Then ad x/2 to both sides. Similarly add y/2. – Cristhian Gz Oct 28 '14 at 13:05

$$a < b$$
Add $a$ on both sides on the one hand, and also $b$ on the other hand:
$$2a < a+b \\ a+b < 2b$$ this gives you $$2a < a+b < 2b \\\implies a<\frac{a+b}2<b$$
Since $a < b$, we have $$a = \frac{a}{2} + \frac{a}{2} < \frac{a}{2} + \frac{b}{2} < \frac{b}{2} + \frac{b}{2} = b.$$
Multiply by 2: $$a<(a+b)/2<b\iff 2a<a+b<2b$$ Now you should clearly see that $2a<a+b\iff a<b\iff a+b<2b$.