In this post, we show that if $latex a$ and $latex b$ are real numbers and $latex a$ is less than $latex b$, then their arithmetic mean is greater than $latex a$. That is, $latex a < b$ implies

$latex a < \displaystyle \frac{a + b}{2}$.

**Proof 1: Direct Proof**

We know that $latex a < b$.

Dividing both sides by 2, we have

$latex \displaystyle \frac{a}{2} < \displaystyle \frac{b}{2}$.

Adding $latex \frac{a}{2}$ to both sides results to

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

which equals

$latex a < \displaystyle \frac{a + b}{2}$.

**Proof 2: Indirect Proof**

Let us assume that for $latex a < b$, $latex a \geq \frac{a + b}{2}$.

Multiplying both sides by $latex 2$ give us $latex 2a \geq a + b$.

Subtracting $latex a$ from both sides results to $latex a \geq b$.

This is in contradiction with the assumption above.

This means that the contradiction is false.

Therefore, $latex a < \displaystyle \frac{a + b}{2}$.