How can I show that for positive reals
$$ \frac{a}{b}\leq \frac{c}{d}, $$ that $$ \frac{a}{b}\leq\frac{a+c}{b+d}\leq\frac{c}{d}. $$
Thanks in advance.
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1Hint : \begin{eqnarray} ab+\color{red}{ad} \leq ab + \color{red}{bc} \ \end{eqnarray} – Donald Splutterwit Dec 02 '20 at 16:06
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1similarly, $ad + cd \leq bc + cd$ – Math Lover Dec 02 '20 at 16:07
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1Does this answer your question? Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$ – Martin R Dec 02 '20 at 16:11
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Also: https://math.stackexchange.com/q/1989104, https://math.stackexchange.com/q/205654, and some more: https://math.stackexchange.com/questions/linked/205654 – Martin R Dec 02 '20 at 16:11
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1@DonaldSplutterwit Thanks, this was a great hint! – Turbotanten Dec 02 '20 at 16:12
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Are you assuming $b,d > 0$. (which is a valid definition for the rationals-- that the denominators must be a positive natural number but the numerator may be any integer). If $b,d$ may be any non-zero integers this needn't be true I think $\frac 1{-2}\le \frac 13$ but $\frac 1{-2} < \frac {1+1}{-2+3}; \frac {1+1}{-2+3} > \frac 13$. – fleablood Dec 02 '20 at 16:20
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@fleablood I forgot to mention that I was assuming all numbers to be positive reals. – Turbotanten Dec 02 '20 at 16:30
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So not merely integers. Cool. I think its true so long as $b,d$ are positive. $a,c$ could be $0$ or negative and it'd still be true. – fleablood Dec 02 '20 at 16:49
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Hint:
$$\dfrac{a+c}{b+d}-\dfrac ab=\dfrac{ab+bc-(ab+ad)}{b(b+d)}=\dfrac{bd\left(\dfrac cd-\dfrac ab\right)}{b(b+d)}$$ which will be $\ge0$ if $\dfrac d{b+d}\ge0$
lab bhattacharjee
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