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 $$ac\leq xy\leq bd.$$
$\endgroup$ 22 Answers
$\begingroup$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.
$\endgroup$ 0 $\begingroup$In generally, if you have
$$a \leq x \leq b$$
$$c \leq y \leq d$$
with $a,b, c, d \in \mathbb{R}$ you have that
$$xy \in [\min(ac, ad, bc), \max(bd, ad, cb)]$$
$\endgroup$ 2
