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.

  • $\begingroup$ Well the max/min of S will occur precisely for the max/min of $2^S = abc$. AM/GM tells us the max is a=b=c=4/3. For min: wolog $a \le b \le c$ and let $a = 1+h, b= 1+k, c = 2 - h - k$. $abc = (1 +h)(1+k)(2-(h+k))$. I think it's inelegant but I think we can show that has a minimum value of 2. $\endgroup$ – fleablood Jul 13 '17 at 16:51

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.


  • $\begingroup$ Nice. By Karamata itself and $(2,1,1) \succ (a,b,c) \succ (\frac43, \frac43, \frac43)$ we have both max and min. +1. $\endgroup$ – Macavity Jul 13 '17 at 18:52
  • $\begingroup$ @Macavity Yes, of course! AM-GM it's Karamata! $\endgroup$ – Michael Rozenberg Jul 13 '17 at 18:59

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


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.

  • 1
    $\begingroup$ When $a=b=c=\frac43$ that is the maximum not the minimum. $\endgroup$ – kingW3 Jul 13 '17 at 17:08
  • $\begingroup$ GM will always be greater than or equal to the HM. Doesn't that mean it is minimum when it is equal to HM? This occurs when $a = b= c$. $\endgroup$ – Agile_Eagle Jul 13 '17 at 17:10
  • $\begingroup$ when $4a=b=c=\frac{4}{3}$ we get the minimum $\endgroup$ – Dr. Sonnhard Graubner Jul 13 '17 at 17:25
  • $\begingroup$ 4/3 + 4/3 + 4/12 = 3 not 4. If b=c= 4/3 then a = 4 - 8/3= 4/3 and that's the max at $3*\log_2 4/3\approx 1.245....$. If $a = b=1;c =2$ the value is $\log_2 1 + \log_2 1 + \log_2 2 = 1$ which is less. $\endgroup$ – fleablood Jul 13 '17 at 20:10

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 $a=b=1; c= 2$.

  • $\begingroup$ So what does your minimum come out to be? I am curious:) As in the approx value, because i dont seem to get your method. My fault though. $\endgroup$ – Agile_Eagle Jul 13 '17 at 19:39
  • $\begingroup$ Um... minimum is $log_2 1 + \log_2 1 + \log_2 2 = 1$. and max is $\log_2 4/3 + \log_2 4/3 + \log_2 4/3 = 3\log_2 4/3 = 3*\sqrt[3]{2}$. $\endgroup$ – fleablood Jul 13 '17 at 19:43
  • $\begingroup$ Note $\log_2 a + \log_2 b + \log_2 c = \log_2 abc$. so for $a=b=1;c=2$ we get $\log_2 2 = 1$ is the min, and $a=b=c=4/3$ and $\log_2 64/27 = 6 - \log_2 27 = 6 - 3\log_2 3 = 3(2- \log_2 3)$ [obvious I made an error above: $\log ab = \log a - \log b$ and not $\sqrt[b]\log a$. D'uh!] $3*\log_2 4/3 \approx 1.245$. $\endgroup$ – fleablood Jul 13 '17 at 19:51
  • $\begingroup$ My method is basically $abc = 2^{\log_2 abc} = 2^{\log a + \log b + \log c} = 2^S$ is increasing or decreasing precisely when $abc$ is increasing or decreasing. And $abc$ where $a+b+c = 4$ achieves a maximum once when $a = b = c$. And $abc$ is least at extreme values of $a,b,$ and $c$. And the extreme values or $a,b,c$ under the conditions $a,b,c \ge 1$ and $a+b+c = 0$ are two of the terms being 1 and the third being 2. The maximum being a=b=c is AM-GM and that $abc$ has one max and monotonic elsewhere I determined by simple calculus. $\endgroup$ – fleablood Jul 13 '17 at 20:01

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