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Simulasi OSN Matematika KTO Mei 2017 No 3

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Diberikan bilangan real positif $a$, $b$, dan $c$ yang memenuhi $a+b+c=1$. Buktikan bahwa \[\frac{c}{\sqrt{ab+1-c}}+\frac{a}{\sqrt{bc+1-a}}+\frac{b}{\sqrt{ca+1-b}} \geq \frac{3}{\sqrt{7}}.\]

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On 28/5/2017 at 9:14 AM, -_- said:
Diberikan bilangan real positif aa , bb , dan cc yang memenuhi a+b+c=1a+b+c=1 . Buktikan bahwa
cab+1c+abc+1a+bca+1b37.cab+1−c+abc+1−a+bca+1−b≥37.
Spoiler

$$(\frac{c}{\sqrt{ab+1-c}}+\frac{a}{\sqrt{bc+1-a}}+\frac{b}{\sqrt{ca+1-b}})^2(3abc+2(ab+bc+ac))\geq 1$$  $$\frac{c}{\sqrt{ab+1-c}}+\frac{a}{\sqrt{bc+1-a}}+\frac{b}{\sqrt{ca+1-b}} \geq \frac{1}{\sqrt{3abc+2(ab+bc+ac)}}$$  Karena $$ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{1}{3}$$ dan $$abc\leq \frac{(a+b+c)^3}{27}=\frac{1}{27}$$  Maka : $$\frac{c}{\sqrt{ab+1-c}}+\frac{a}{\sqrt{bc+1-a}}+\frac{b}{\sqrt{ca+1-b}} \geq \frac{1}{\sqrt{3abc+2(ab+bc+ac)}}\geq \frac{3}{\sqrt{7}}$$  Done :)

$CMIIW$

Edited by username
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On Wednesday, May 31, 2017 at 10:14 PM, username said:
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(cab+1c+abc+1a+bca+1b)2(3abc+2(ab+bc+ac))1(cab+1−c+abc+1−a+bca+1−b)2(3abc+2(ab+bc+ac))≥1

 

cab+1c+abc+1a+bca+1b13abc+2(ab+bc+ac)cab+1−c+abc+1−a+bca+1−b≥13abc+2(ab+bc+ac)

 Karena

ab+bc+ac(a+b+c)23=13ab+bc+ac≤(a+b+c)23=13

dan 

abc(a+b+c)327=127abc≤(a+b+c)327=127

 Maka :

cab+1c+abc+1a+bca+1b13abc+2(ab+bc+ac)37cab+1−c+abc+1−a+bca+1−b≥13abc+2(ab+bc+ac)≥37

 Done :)

 

CMIIWCMIIW

 

cab+1−c−−−−−−−−√+abc+1−a−−−−−−−−√+bca+1−b−−−−−−−−√)2(3abc+2(ab+bc+ac)) kok bisa ≥ 1 ya??

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