Inequality : definition interval We have the following parameters :
$y \in [0,1]$
$z \in [0,1]$
and I have now to proof the following thing:
$(y+z-2yz) \in [0,1]$
$0 \leq y \leq 1$
$0 \leq z \leq 1$
Therefore  $\Rightarrow 0 \leq y + z \leq 2 \tag{1}$
Then, $0 \leq yz \leq 1$ $\Rightarrow 0 \leq 2yz \leq 2 $ $\Rightarrow -2 \leq - 2yz \leq 0 \tag{2}$
But $(1) + (2)$ :
$-2 \leq y + z - 2yz \leq 2 $
and not $0 \leq y + z - 2yz \leq 1 $.
 A: All of steps taken are valid. However, you are not interpreting the result in the correct way.
See for example: Let $x\in (0,1)$, so $0<x<1$, then we have $-5<x<5$, or $x\in(-5,5)$. Thus, the possible values of $x$ are in $(-5,5)$; However, note that it is not true that every value in $(-5,5)$ is actually attained. This is due to the fact that $-5$, $5$ are not the "optimal" bounds for $x$.
Comparing to your result, it is true that, by your deduction, $-2\leq y+z-2yz\leq 2$. However, your method of considering the bounds of each term individually is not "sharp" enough that it does not give the optimal bounds you're looking for, and hence does not prove your assertion. More precisely, the reason why your method is not sharp enough is the upper (and lower, resp.) bounds of $y+z$ and $-2yz$ are not attained at the same value of $(y,z)$. Hence, the sum of upper (and lower, resp.) bounds is not an actual attainable upper (and lower, resp.) bound.
The "sharper" approach that gives the optimal bound and also proves your assertion would take into account terms together as follows:
Let $y,z\in [0,1]$. It is easy to see that $y-\frac{1}{2}\in \left[-\frac{1}{2},\frac{1}{2}\right]$ and $1-2z\in [-1,1]$, thus we have that $\left(y-\frac{1}{2}\right)(1-2z) \in \left[-\frac{1}{2},\frac{1}{2}\right]$. Now consider the expression $$y+z-2yz=y+z-2yz-\frac{1}{2}+\frac{1}{2}\\= \left(y-\frac{1}{2}\right)(1-2z)+\frac{1}{2}.$$ Thus, the expression above lies in $\left[-\frac{1}{2}+\frac{1}{2},\frac{1}{2}+\frac{1}{2}\right]=[0,1]$, as desired.
A: Your original method is correct, but is crude because when $y$ increases for example, the first inequality increases but the second decreases.
This solution uses differentiation to find the functions minima and maxima.
Let $y+z=k$, then the function becomes:
$$y+(k-y)-2y(k-y)$$
$$=k-2ky+2y^2$$
Treat $k$ as a constant, and differentiate wrt $y$:
$$-2k+4y$$
This is zero when $y=\frac{k}{2}$, and this is a minimum from the second differential.
Similarly for $z$, and so the line $y=z$ contains the minimum value for the function, which is $0$.
As the original function is a quadratic in $y$, the maximum is obtained on the border of the domain, and on the border the function is linear, hence the maximum value is $1$.
