Let $a, b, c$ be positive real numbers such that $a b c=8$. Find the minimum value that the expression $a+b+c$ can take.
The form of this exercise is a bit different from the previous ones, but we should not be destabilized by it. To find the minimum value that $a+b+c$ can take, we need to lower-bound this expression, that is, find an inequality of the form $a+b+c \geqslant x$ where $x$ will be a real number, and then find a case of equality for this inequality (since we are looking for the minimum value, an inequality alone is not enough). Given that we have a hypothesis on the product, it is natural to apply the AM-GM inequality to see what it yields. We get $a+b+c \geqslant 3 \sqrt[3]{a b c}=3 \cdot 2=6$. Be careful, this does not suffice to conclude that 6 is the minimum value; our inequality might be too crude, so we still need to show that this value can be achieved! Fortunately, we know that the AM-GM inequality has an equality case when $a=b=c$. This equality can be achieved while still satisfying the hypothesis $a b c=8$ by choosing $a=b=c=2$, and we verify that we indeed get $a+b+c=2+2+2=6$, so 6 is indeed the minimum value sought.