Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$

Question

Can anyone help me solve the two cases that are derived from this problem?

try to find at least one pair of real numbers, for which the inequality doesn't work... — V-X, May 31 '13 at 10:41
@V-X: The key, there, is that one is trying to derive a contradiction, yes? If one doesn't keep that in mind, such an approach is a waste of time. — Cameron Buie, May 31 '13 at 10:53

score 8 · Answer 1 · answered May 31 '13 at 10:42

8

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.

answered May 31 '13 at 10:42

Ittay Weiss

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score 5 · Answer 2 · answered May 31 '13 at 14:26

5

$$ \begin{eqnarray} \text{min}(a, b)&\le& a\\ \text{min}(a, b)&\le& b\\ 2 \text{min}(a, b)&\le& a + b\\ \text{min}(a, b)&\le& \frac{a + b}{2}. \end{eqnarray} $$

answered May 31 '13 at 14:26

Eric Jablow

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score 2 · Answer 3 · answered May 31 '13 at 10:39

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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, $$\min\{a,b\}=\frac{a+b}2-\frac{|a-b|}2.$$

answered May 31 '13 at 10:39

Cameron Buie

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My issue is with assuming that a ≤ b and proving the case. I don't know how to proceed. – saurs May 31 '13 at 10:40
@saurs You have $2$ numbers, $x$ and $y$. Define the smaller of the two to be $a$, and the larger to be $b$. – Meow May 31 '13 at 10:44
@saurs: If you've assumed that $a\le b$, then $\min{a,b}=???$ – Cameron Buie May 31 '13 at 10:44
min{a,b} = a. I see your point. – saurs May 31 '13 at 10:48
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A user just deleted their answer but I was wondering if it can be considered a good approach because of its simplicity? – saurs May 31 '13 at 10:58
@saurs: Certainly. In fact, that's pretty much what I'd recommend. – Cameron Buie May 31 '13 at 11:03

score 2 · Answer 4 · answered May 31 '13 at 10:47

2

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 $B\ge 0$.

answered May 31 '13 at 10:47

egreg

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DonAntonio · Accepted Answer · 2013-05-31T16:06:29.617

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Hint:

$$\forall\,a\,,\,b\in\Bbb R\;,\;a<b\implies\frac{a+b}2>\frac{a+a}2\;\ldots$$

edited May 31 '13 at 16:06

answered May 31 '13 at 15:58

DonAntonio

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score 0 · Answer 6 · answered May 31 '13 at 10:50

0

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.

answered May 31 '13 at 10:50

Riccardo.Alestra

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score 0 · Answer 7 · answered May 31 '13 at 15:10

0

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.

answered May 31 '13 at 15:10

alexis

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Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$

7 Answers7