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

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  • $\begingroup$ try to find at least one pair of real numbers, for which the inequality doesn't work... $\endgroup$
    – V-X
    May 31, 2013 at 10:41
  • $\begingroup$ @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. $\endgroup$ May 31, 2013 at 10:53

7 Answers 7

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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.

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

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

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  • $\begingroup$ My issue is with assuming that a ≤ b and proving the case. I don't know how to proceed. $\endgroup$
    – saurs
    May 31, 2013 at 10:40
  • $\begingroup$ @saurs You have $2$ numbers, $x$ and $y$. Define the smaller of the two to be $a$, and the larger to be $b$. $\endgroup$
    – Meow
    May 31, 2013 at 10:44
  • $\begingroup$ @saurs: If you've assumed that $a\le b$, then $\min\{a,b\}=???$ $\endgroup$ May 31, 2013 at 10:44
  • $\begingroup$ min{a,b} = a. I see your point. $\endgroup$
    – saurs
    May 31, 2013 at 10:48
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    $\begingroup$ A user just deleted their answer but I was wondering if it can be considered a good approach because of its simplicity? $\endgroup$
    – saurs
    May 31, 2013 at 10:58
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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 $B\ge 0$.

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Hint:

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

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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.

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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.

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