Is the minimum of the product of two functions equal to the product of their minima? I have stuck with following equality, 
For all $x$, assume function $a(x)$, $b(x)$ have nonzero, and non negative values. (i.e $a(x)>0$,  $b(x)>0$, 
Is the following equality true?
$$\min(ab)=\min(a)\min(b)$$
From the below answers, i need more constraint to achieve, this equality. 
How about function $a(x)$, $b(x)$ are $C^{n}$  ($n$th differentiable, i.e smooth function) 
then is this equality holds?  (Is the condition for differentiablity is necessary? )

It seems plausible to change this problem into set theory language, 
For set $A$ and $B$ with nonzero non-negative elements, 
assume element $a \in A$, $b \in B$, is this equality true?
$$\min(ab)=\min(a)\min(b)$$
I tried to look up some analysis textbook, but I could not find satisfactory proofs.
 A: It's only true if $a(x)$ and $b(x)$ attain their minimum for the same value of $x$. 
For instance, consider $a(x) = (x+1)^2+1$ and $b(x) = (x-1)^2+1$. 
Clearly, $\min a = \min b = 1$. But $a(x)b(x) = x^4+4$, for which $\min ab = 4$. 

EDIT: The OP recently added the following to the question: 

From the below answers, i need more constraint to achieve, this
  equality.  How about function $a(x)$, $b(x)$ are $C^{n}$  ($n$th
  differentiable, i.e smooth function)  then is this equality holds? 
  (Is the condition for differentiablity is necessary? )

Adding the constraint that $a(x)$ and $b(x)$ are $C^n$ functions, or even $C^{\infty}$ functions, or even analytic functions won't help, as my above counterexample still meets these additional constraints.
Note that under the given conditions, we can prove that $\min(ab) \ge \min(a)\min(b)$. But as I stated before, equality only holds if there is an $x_0$ such that $a(x_0) = \min(a)$ and $b(x_0) = \min(b)$. 
A: Hint
Just consider $$\Big(a(x)b(x)\Big)'=a'(x)b(x)+a(x)b'(x)$$ If it exists $x_0$ such that $a'(x_0)=b'(x_0)=0$ then,for $x=x_0$, $\Big(a(x)b(x)\Big)'=0$. This is the required condition.
A: It is not true. Consider the two functions:
$$f(x)=1+|x-1|,\qquad g(x)=1+|x+1|.$$
Obviously $\min f=\min g=1$, but $\min(fg)=3$.
A: it's not correct. for example assume $a(x)=x^2 $ and $b(x)=1/x^2$ 
$min(a)=0$ and $b$ doesn't have minimum but $min(ab)=min(1)=1$
