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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.

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  • $\begingroup$ 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
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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!

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  • $\begingroup$ 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
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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.

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

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  • $\begingroup$ 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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