If $p,q,r$ are positive real numbers and $p+ q+ r = \frac{1}{2}$ , then find the range of $(1-p)(1-q)(1-r)$ 
If $p,q,r$ are positive real numbers and $p+q+r = \frac{1}{2}$, then find the range of $(1-p)(1-q)(1-r)$. 

Here, one can easily reach the upper limit, but what about the lower limit? Somebody please help me with this problem. 
 A: We’ll at first permit that $x$, $y$, $z$ are zero. We’ll discard these cases at the end.
The set $$S=\left\{(x,y,z)\in\left(\mathbb R^+_0\right)^3\mid x+y+z=\frac12\right\}$$ is closed, and $$f:S\to\mathbb R;\,f(x,y,z)=(1-x)(1-y)(1-z)$$ is continuous. Therefore, $f$ attains both a maximum and a minimum.
To find them, we can simply use that, for $x$, $y$ with a constant sum $k$, $$(1-x)(1-y)=1-k+xy=(1-k)+x(k-x)\\=(1-k)-\left(x-\frac k2\right)^2+\frac{k^2}4=\left(1-k+\frac{k^2}4\right)-\frac14\left|x-y\right|^2 $$ is greater when $x$ and $y$ are closer together, and smaller when the opposite happens.
Therefore, we can conclude that the maximum of $f$ is attained at a point $(x,y,z)$ where the pairwise differences can’t all be simultaneously further decreased, i.e. $\left(\frac16,\frac16,\frac16\right)$, and that its minimum occurs at a point where they can’t be simultaneously further increased, i.e. $(x,y,0)$, or some permutation thereof. In this case, $$f(x,y,0)=(1-x)(1-y)$$ is easily seen to be minimized when $x=\frac12$ and $y=0$, or viceversa. In particular, if we restrict $x$, $y$, $z$ to the positive reals, we’re unable to get the minimum value, but we can get arbitrarily close, taking $\left(\frac12-2\varepsilon,\varepsilon,\varepsilon\right)$.
By plugging in these numbers, we get that the range of our original expression is $$\boxed{\left(\frac12,\frac{125}{216}\right]}.$$ $\blacksquare$
A: With $x=1-p,\>y=1-q,\>z=1-r$, The problem is equivalent to: 

Find the range of $xyz$, subject to $x+y+z=\frac52$ and $x,y,z<1$.

Reexpress  $ xyz= f(x,y)  = xy(\frac52-x-y)$ and find the extrema of $f_x'=f_y'=0$, which leads to 
$$\frac52-x-2y=\frac52-y-2x=0\implies x=y = \frac56$$
It can be verified that the resulting $f(\frac56,\frac56) = \frac{125}{216} $ is the maximal. For the minimum, examine $f(x,y)$ along the boundaries,
$$f(x,1) = x(\frac32-x) \le \frac9{16},\>\>\>\>\>f(1,y) 
= y(\frac32-y) \le \frac9{16},\>\>\>\>\>
f(1,1)=\frac12$$
Thus, the minimum is $\frac12$. 
