Minimum of product of 2 functions

In summary, if both functions have nonnegative values in the domain, then the product of the minima is less than the minimum of products.
  • #1
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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
 
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  • #2
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 $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,
 
  • #3
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 $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,

I am extremely grateful
Sarrah
 

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