MHB Minimum of product of 2 functions

[sarrah1]

Hello

Simple question

Whether the minimum of the product of two functions in one single variable, is it greater or less than the product of their minimum
thanks
Sarrah

 

[Evgeny.Makarov]

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 $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,

 

[sarrah1]

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 $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,

I am extremely grateful
Sarrah