Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$ Can anyone help me solve the two cases that are derived from this problem?
 A: Assume $m = \min\{a,b\}> \frac{a+b}{2}$. Then $m+m>\frac{a+b}{2}+\frac{a+b}{2}=a+b\ge m+m$, a contradiction. 
A: $$
\begin{eqnarray}
\text{min}(a, b)&\le& a\\
\text{min}(a, b)&\le& b\\
2 \text{min}(a, b)&\le& a + b\\
\text{min}(a, b)&\le& \frac{a + b}{2}.
\end{eqnarray}
$$
A: Hint: You may as well assume that $a\le b$. (Why is that okay?)
In fact, you should be able to show that $\min\{a,b\}$ is the midpoint of $a,b$ minus half the distance between $a,b$--that is, $$\min\{a,b\}=\frac{a+b}2-\frac{|a-b|}2.$$
A: Let $c=\min\{a,b\}$ and set $A=a-c$, $B=b-c$; observe that $A\ge0$ and $B\ge0$. Then $a=A+c$, $b=B+c$, so
$$
\frac{a+b}{2}=\frac{A+c+B+c}{2}=c+\frac{A+B}{2}\ge c
$$
because $A\ge 0$ and $B\ge 0$.
A: Hint:
$$\forall\,a\,,\,b\in\Bbb R\;,\;a<b\implies\frac{a+b}2>\frac{a+a}2\;\ldots$$
A: If $a=b$ you get:
$$min\{a,a\}=\frac{1}{2}2a=a$$
If $a>b$,
$$min\{a,b\}=b$$
So
$$a+b\gt 2b$$
and $a-b\gt0$. Because $a\gt b$ this is true.
A: Just multiply both sides by two: 
$$2 \cdot \min(a, b) < a + b$$
This is necessarily true since either $a$ or $b$ is greater than $\min(a,b)$, and the other is equal to it.
