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For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.

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In addition to thanking the answerers, you should vote up all of the answers you find useful. Click on the grey arrow pointing up. – Byron Schmuland Jun 23 '12 at 19:20
up vote 1 down vote accepted

Either $\frac{1}{2a}\leq \frac{1}{2b}$, in which case $$\frac{1}{b}=\frac{1}{2b}+\frac{1}{2b}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2a}+\frac{1}{2a}=\frac{1}{a},$$ or $\frac{1}{2b}\leq \frac{1}{2a}$, in which case $$\frac{1}{a}=\frac{1}{2a}+\frac{1}{2a}\geq\underbrace{\frac{1}{2a}+\frac{1}{2b}}_{\atop \dfrac{1}{c}}\geq\frac{1}{2b}+\frac{1}{2b}=\frac{1}{b}.$$ Now take reciprocals.

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Thanks a lot.11 – Miau Jun 23 '12 at 19:08

In general, for any $x_1,x_2,\ldots,x_n$, if we have weights $w_1,w_2,\ldots,w_n \in [0,1]$ such that $\displaystyle \sum_k w_k = 1$, then the weighted average i.e. $$\sum_k w_k x_k \in [\min_j \{x_j\}, \max_j \{x_j\}]$$ The above is true since $$\min_j \{x_j\} \leq x_k \leq \max_j \{x_j\} \implies \sum_k w_k \min_j \{x_j\} \leq \sum_k w_k x_k \leq \sum w_k \max_j \{x_j\}\\ \implies \min_j \{x_j\} \leq \sum_k w_k x_k \leq \max_j \{x_j\} \left(\text{Since } \sum_k w_k = 1 \right)$$ In your case, $x_1 = \dfrac1a$, $x_2 = \dfrac1b$ and the weights are $w_1 = w_2 = \dfrac12$. Hence, we get that $$\dfrac1c \in \left[\min \left(\dfrac1a, \dfrac1b \right), \max \left(\dfrac1a, \dfrac1b \right) \right] \implies c \in [\min(a,b),\max(a,b)]$$

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Thanks a lot.!! – Miau Jun 23 '12 at 19:08

Hint $\ $ For $\rm\ A = 1/a,\ B = 1/b,\ C = 1/c\ $ it becomes obvious, viz. $$\rm C = \frac{A+B}2\ \Rightarrow\ A \ge C \ge B\ \ or\ \ B \ge C \ge A$$

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Thanks a lot~~1 – Miau Jun 23 '12 at 19:31

$c\geq a\Rightarrow 1/2a+1/2b\leq 1/a\Rightarrow1/2b\leq1/2a\Rightarrow 1/2a+1/2b\geq1/b\Rightarrow c\leq b$. Similarly do when $c\leq a$

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Thanks a lot.!! – Miau Jun 23 '12 at 19:08
@Bill Dubuque: This becomes obvious because we are more familier with this but the proof requires exactly same idea, this is funny! – pritam Jun 23 '12 at 19:33

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