The minimum of the product of two nonnegative functions is always greater than or equal to the product of their minima. This is established by the property of real numbers that states if both functions yield nonnegative values, then the inequality holds: if $0 \leq x_1 \leq y_1$ and $0 \leq x_2 \leq y_2$, then $x_1 x_2 \leq y_1 y_2$. However, if either function can yield negative values, this relationship does not hold, as demonstrated by the example where $a_1=1$, $a_2=2$, and $b_1=b_2=-1$. In this case, the product of minima exceeds the minimum of products.
PREREQUISITESMathematicians, students of calculus, and anyone interested in optimization and function analysis will benefit from this discussion.
Evgeny.Makarov said:If the values of both functions are nonnegative, then the product of minima is less than or equal to the minimum of products. This relies on the following fact about real numbers: if $0\le x_1\le y_1$ and $0\le x_2\le y_2$, then $x_1x_2\le y_1y_2$.
For example, consider functions $f$ and $g$ and just two values in the domain: $x_1$ and $x_2$. Let $a_1=f(x_1)$, $a_2=f(x_2)$, $b_1=g(x_1)$, $b_2=g(x_2)$, Then $\min(a_1,a_2)\le a_1$ and $\min(b_1,b_2)\le b_1$, so by the fact above $\min(a_1,a_2)\min(b_1,b_2)\le a_1b_1$. Similarly $\min(a_1,a_2)\min(b_1,b_2)\le a_2b_2$, so $\min(a_1,a_2)\min(b_1,b_2)\le\min(a_1b_1,a_2b_2)$.
If the numbers can be negative, then this conclusion no longer holds. For example, if $a_1=1$, $a_2=2$ and $b_1=b_2=-1$, then $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,