# True or false: " the minimum value of a product is the product of the minimum value of its factors"? Under which conditions?

[Edited]

The sentence " the minimum value of a product is the product of the minimum values of its factors" seems comformable to common sense.

Is it correct however? What qualifications would it require?

[ Previous version, involving calculation mistakes]

The sentence " the minimum value of a product is the product of the minimum values of its factors" seems comformable to common sense.

However, while thinking about the question ( asked recently) : " what is the set of values the expression $$2x^2 - 2x$$ can take while $$x$$ ranges over $$[0,2]$$?" , the above intuitive principle induced me into error.

The original expression can be factored : $$2x^2 - 2x$$ = $$(2x) (x-1)$$

If $$x$$ ranges over $$[0,2]$$, then the minimum value of $$2x$$ is $$0$$ and the minimum value of $$x-1$$ is $$-1$$. .

So, apparently, the minimum value of the product should be $$0$$ ( that is, the product of the minimum values of the factors).

But in fact, the minimum value is $$-\frac 12$$ ( as a graph shows).

Hence my question : how to show that the alledged intuitive principle is wrong? or what qualifications would it require?

• The minimal value of $x+1$ is $1$ not $-1$ and the minimal value of the product is $0$ and not $-1$: If $x \geq 0$, then $2x^2 + 2x \geq 0$, but for $x =0$ we obtain $0$ which therefore is the minimal value.
– Con
Mar 15, 2020 at 11:16
• Thanks for your comment, I'll edit.
– user655689
Mar 15, 2020 at 11:18
• Please stop deleting your post, it's rather disruptive. If what you mean by "minimum value of factor" is its minimum anywhere, then a nice straightforward counterexample is $(x + 1)(2 - x)$ over $[0, 1]$. Some sufficient qualifications would be "if the minimum absolute values of the factors occur at the same point, then the minimum absolute value of the product is also at that point". Mar 15, 2020 at 11:22
• The minimum of $x^2+2x$ over $[0, 2]$ is not $-1/2$. Mar 15, 2020 at 11:26

• What astonishes me is that an analogous principle ( involving maximum value of a product) is used in $\delta$ $-$ $\epsilon$ proofs ( or rather in the preliminary work aiming at finding a suitable $\delta$ value.
• There is an inequality like the one used to prepare $\epsilon$--$\delta$ proofs. $$\min\alpha(t)\beta(t) \ge \big(\min \alpha(t)\big)\big( \min \beta(t)\big)$$when $\alpha(t), \beta(t) \ge 0$. The inequality with $\max$ has $\le$ in place of $\ge$ Mar 15, 2020 at 13:11