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2130 posts

Sayan

#1 h May 12, 2013, 11:04 AM • 3 Y

Y by Samujjal101, Adventure10, Mango247

Let $a,b,c$ be real number greater than $1$ . Let
$S=\log_a {bc}+\log_b {ca}+\log_c {ab}$
Find the minimum possible value of $S$ .

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15142 posts

#2 h May 12, 2013, 11:18 AM • 2 Y

Y by Adventure10, ThEdArK0

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2130 posts

Sayan

#3 h May 12, 2013, 2:36 PM • 3 Y

Y by hemangsarkar, Adventure10, Mango247

My solution in exam (almost identical to mavropnevma):
$S+3=\log_a {abc}+\log_b {abc}+\log_c {abc}$
Let
$T=\log_{abc} a+\log_{abc} b+\log_{abc} c =1$
Observe by Cauchy schwarz:
$T(S+3) \ge (1+1+1)^2 =9 \implies S\ge 6$

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Potla

#4 h May 12, 2013, 2:58 PM • 2 Y

Y by Adventure10 and 1 other user

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#5 h May 13, 2013, 9:28 AM • 2 Y

Y by ac9991, Adventure10

Sayan wrote:

Let $a,b,c$ be real number greater than $1$ . Let
$S=\log_a {bc}+\log_b {ca}+\log_c {ab}$
Find the minimum possible value of $S$ .

Write it as ,
$S = \frac{\log{b}}{\log{a}} + \frac{\log{c}}{\log{a}}+\frac{\log{c}}{\log{b}}+\frac{\log{a}}{\log{b}} + \frac{\log{a}}{\log{c}}+\frac{\log{b}}{\log{c}} \stackrel{AM-GM}{\ge} 6 \Box$

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#6 h May 29, 2013, 3:06 PM • 1 Y

Y by Adventure10

As a,b,c>1 $log_a{bc},log_b{ac},log_c{ab}\ge0$ then: $S=\log_a{b}+\log_a{b}+\log_b{a}+\log_b{c}+\log_c{a}+\log_c{b}$ then:
$M=\log_a{b}+\log_b{a}\ge2$
$N=\log_b{c}+\log_c{b}\ge2$
$K=\log_c{a}+\log_a{c}\ge2$
then:
$M+N+K\ge6$

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579 posts

#7 h Mar 26, 2015, 6:40 AM • 3 Y

Y by stranger_02, Adventure10, Mango247

This inequality is easy just you need to know property of log and AM-GM.
$S= \log_a {bc}+\log_b {ca}+\log_c {ab} = \frac{\log{bc}}{\log{a}}+\frac{\log{ca}}{\log{b}}+\frac{\log{ab}}{\log{c}}$
which on simplification gives,
$\frac{\log{b}}{\log{a}}+\frac{\log{c}}{\log{a}}+\frac{\log{c}}{\log{b}}+\frac{\log{a}}{\log{b}}+\frac{\log{a}}{\log{c}}+\frac{\log{b}}{\log{c}}]$
now we apply the famous AM-GM inequality, which gives $S\ge 6$

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R-sk

#9 h Feb 7, 2021, 8:54 PM

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Trivial just apply am gm by using change the base formula for logariths

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#10 h Feb 6, 2022, 7:54 AM

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We can apply am gm here as a, b, c all belongs to positive reals>1. So by am gm S>=6. Smallest S=6,when equality holds i. e. a=b=c

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lifeismathematics

1188 posts

lifeismathematics

#11 h May 26, 2022, 7:03 PM

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good one

This post has been edited 1 time. Last edited by lifeismathematics, May 26, 2022, 7:03 PM

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#12 h Nov 2, 2023, 1:24 PM

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lifeismathematics wrote:

good one

Let $\log_a {bc}=x$ , $\log_b {ca}=y$ , $\log_c {ab}=z$ be their minimum value
this implies,
$a^x=bc$
$b^y=ca$
$c^z=ab$
Multiplying all three gives
$a^x b^y c^z=a^2b^2c^2$

IF we compare ?
values of $x,y,z=2$
so $S_{min}=\boxed{6}$

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#13 h Mar 8, 2025, 3:36 PM

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solution

This post has been edited 1 time. Last edited by quasar_lord, Mar 8, 2025, 4:01 PM

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#14 h Mar 8, 2025, 8:59 PM

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Storage

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#15 h Wednesday at 6:24 PM

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Simple application of AM-GM
Since they are all real numbers greater than 1 ,so we can apply AM-GM and we get S>=6
Hence we obtain the minimum value of S is 6 , which is achievable when a=b=c.

This post has been edited 2 times. Last edited by Apple_maths60, Wednesday at 6:26 PM
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