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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

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    $\begingroup$ from x<y you know x/2 < y/2. Then ad x/2 to both sides. Similarly add y/2. $\endgroup$ – Cristhian Gz Oct 28 '14 at 13:05
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$$ a < b $$

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 $$ 2a < a+b < 2b \\\implies a<\frac{a+b}2<b $$

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Since $a < b$, we have $$a = \frac{a}{2} + \frac{a}{2} < \frac{a}{2} + \frac{b}{2} < \frac{b}{2} + \frac{b}{2} = b.$$

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Multiply by 2: $$a<(a+b)/2<b\iff 2a<a+b<2b$$ Now you should clearly see that $2a<a+b\iff a<b\iff a+b<2b$.

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