Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$. Let $a, b, c \geq 1$ and $a+b+c=4$. Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$.
I found the maximum, it is easy to prove $S_\max = 3\log_{2}\frac{4}{3}$. I think the minimum is $1$ when there are two number are $1$ and the remain is $2$. But I do not know, how to prove it.
 A: Let $f(x)=\log_2{x}$ and $a\geq b\geq c$. Hence, $f$ is a concave function.
Also we have: $a\leq2$ and $a+b=4-c\leq3$.
Thus, $(2,1,1)\succ(a,b,c)$ and by Karamata 
$$\sum_{cyc}\log_2a\geq\log_22+\log_21+\log_21=1.$$
The equality occurs for $a=2$ and $b=c=1$, which says that $1$ is a minimal value.
In another hand, by AM-GM
$$\sum_{cyc}\log_2a=\log_2abc\leq\log_2\left(\frac{a+b+c}{3}\right)^3=3\log_2\frac{4}{3}.$$
The equality occurs for $a=b=c=\frac{4}{3}$, which says that $\log_2\frac{4}{3}$ is a maximal value.
Done!
A: Basically you want to find minimum of $\log_{2}{abc}$
It follows that this happens when the quantity $abc$ is itself minimum under the constraint $a+b+c=4$ and $a, b, c \geq 1$.
The GM-HM inequality gives,
$$
(abc)^\frac{1}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}
$$
Equality occurs only when $a$, $b$ and $c$ are equal to $\frac{4}{3}$.
This gives the minimum value of $\log_{2}{abc}$ to be 

$$log_{2}{\frac{27}{64}} $$

which is approximately equal to the numerical value 

$$-1.24511249783653145563878316815655047372055677692255681863....$$

As for finding the maximum of the expression, you said you were able to do that easily. One comment mentions that it can be done via the AM-GM inequality.
A: The max/min of $S$ will occur on the same instance as the max/min of $2^S = abc$.  By $AM-GM$ the max occurs at $a= b=c = 4/3$ as you figured.  
To figure out the min, fix $c$ and and set $b = (4-c) -a = K -a$.
So $abc = aK - a^2$ the derivative is $K - 2a$ which has a maximum at $a = \frac K2$ (which means $b = a = \frac K 2$) and will be increasing for $a < \frac K 2$ and decreasing for $a > \frac K2$ so the minimum will occur at either the minimum possible value of $a$ which is $1$ (and therefore $a = 1; b= K-1$) or at the maximum possible value of $a$ which is $K - 1$ and (therefore $a= K-1; b = 1$).  Wolog we may assume $a \le b$ and $a = 1$ and $b = K-1$.
So lets "unfix" $c$ but fix $a$ so as to find min of $abc$ which means fixing $a = 1$. We have $a = 1; b = 4- c - 1= 3-c$ and so $abc = 3c - c^2$.  By a similar argument above, This achieves a maximum at $c = \frac 32$ and minimum at either $c = 1, b=2$ or at $c= 2, b= 1$.
So if, wolog, $a \le b \le c$ then max occurs at $a= b= c = \frac 43$ and min occurs at $a=b=1; c= 2$.
