How would you prove that $m

Question

Let $M,m,a,b\in\Bbb R$. How would you prove that $m

PS: By triangle inequality, we can show that $|a-b|\leq|a-m|+|m-M|+|M-b|$, but this does not show that $|a-b|

Answer 1

Inequalities of real numbers like the one in your question follow easily from the properties of order relations of real numbers, but they are not obvious / self evident unless visualized on the number line. Also note that using number line is rigorous enough because of the assumption that to each point on the number line corresponds a unique real number and vice versa and the linear order from left to right on the number line is equivalent to the ordering of real numbers.

But in case you wish to avoid the number line here is a simple argument. Without any loss of generality we can assume $ab$ can be obtained by reversing the roles of $a, b$. We then have $$mb-a=|a-b|$. The technique of splitting an expression into multiple terms by adding and subtracting the same quantity is heavily used in analysis mainly to establish inequalities like these.

Answer 2

We have $$a - b < M - m$$ and $$-(M - m) = m - M < a - b$$ Hence together $$|a - b| < M-m$$ By using that $$|x| < c \quad \Leftrightarrow \quad -c < x < c$$ for $x \in \mathbb{R}$ and $c > 0$.

Answer 3

Pf I: (A super easy proof)

Note that $\begin{cases}m

Pf II: No matter $a$ or $b$ is the bigger, since $a

Add these two expression, we have $a+(-b)

Since the choice of $a,b$ is arbitrary, then $b-a

Hence no matter $|a-b|=a-b$ or $b-a$, we have $|a-b|