Multiplying two inequalities

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 $$ac\leq xy\leq bd.$$

Solution 1:

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.