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Suppose we have two inequalities $$a\leq x\leq b\tag{1}$$ $$c\leq y\leq d\tag{2},$$ where $a,b,c,d>0$. Then can I conclude that $$ac\leq xy\leq bd\quad ?$$

My attempt: Since $a,b,c,d>0$ and $\log_e$ is monotonic then we can write $$\log a\leq \log x\leq\log b\tag{3}$$ $$\log c\leq \log y\leq\log d\tag{4}$$ Adding $(3)$ and $(4)$, $$\log a+\log c\leq \log x + \log y\leq \log b+\log d.$$ Combining logs gives $$\log(ac)\leq \log(xy)\leq \log(bd).$$ Exponentiating then gives $$ac\leq xy\leq bd.$$

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    $\begingroup$ From a fundamental analysis point of view: If $a,b,c,d$ are all real and all positive and your equations 1, 2 are assumed then according to the properties of $\mathbb{R}$ being an ordered field, along with transitivity that comes from the defining notions of a linear ordering, your conclusion is automatically valid. This is definition 1.17 of Rudin and the following proposition 1.18. $\endgroup$ – Sudarsan May 27 '15 at 19:04
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    $\begingroup$ You can also use that $\frac{b}{x} \geq 1$, $\frac{d}{y} \geq 1$ and that multiplying numbers $\geq 1$ leads to number that is also $\geq 1$ without resorting to logarithms. $\endgroup$ – Sil Dec 15 '18 at 11:09
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Yes your proof is correct. Excellent work reducing the question about multiplying inequalities to a more familiar one of adding inequalities. The only thing I would mention is that taking logarithms and exponentiating are monotone increasing operations. If they were monotone decreasing, the inequalities would flip.

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