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.

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!
