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KTO Matematika

Kontes Terbuka Olimpiade Matematika - Desember 2016 - Bagian B Nomor 4

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Buktikan bahwa untuk semua bilangan real positif $a$, $b$, dan $c$ yang memenuhi $abc \geq 1$ berlaku ketakasamaan

 

\[\frac{1}{a^3+b^2+c^2}+\frac{1}{b^3+c^2+a^2}+\frac{1}{c^3+a^2+b^2} \leq 1.\]

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perhatikan

1>=1/(a^3+b^2+c^2) + 1/(b^3+a^2+c^2) + 1/(c^3+a^2+b^2)>=3^2/(a^3+b^3+c^3+2(a^2+b^2+c^2)) , sehingga sama saja dengan kita membuktikan

1>=3^2/(a^3+b^3+c^3+2(a^2+b^2+c^2)) 

(a^3+b^3+c^3+2(a^2+b^2+c^2))>=3^2

karena abc>=1 , dengan menggunakan am>=gm , terbukti bahwa

(a^3+b^3+c^3+2(a^2+b^2+c^2)) >= 3^2=9

kesamaan terjadi disaat a=b=c=1

 

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On 12/29/2016 at 0:52 PM, steffen123 said:

perhatikan

1>=1/(a^3+b^2+c^2) + 1/(b^3+a^2+c^2) + 1/(c^3+a^2+b^2)>=3^2/(a^3+b^3+c^3+2(a^2+b^2+c^2)) , sehingga sama saja dengan kita membuktikan

1>=3^2/(a^3+b^3+c^3+2(a^2+b^2+c^2)) 

(a^3+b^3+c^3+2(a^2+b^2+c^2))>=3^2

karena abc>=1 , dengan menggunakan am>=gm , terbukti bahwa

(a^3+b^3+c^3+2(a^2+b^2+c^2)) >= 3^2=9

kesamaan terjadi disaat a=b=c=1

 

Hello

Dalam pembuktian Anda, Anda menunjukkan bahwa X>Z, dengan menggunakan fakta bahwa Y>Z. Namun soal meminta Anda untuk membuktikan X>Y. 

Salam Harambe

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Spoiler

$a, b, c \in \mathbb{R}^{+} , abc \geq 1 \Rightarrow \frac{1}{a^3+b^2+c^2}+\frac{1}{b^3+c^2+a^2}+\frac{1}{c^3+a^2+b^2} \leq 1$

  • By $AM \geq GM$

${a+b+c} \geq 3 \sqrt[3]{abc}$

$\Leftrightarrow {a+b+c} \geq 3$

$\Leftrightarrow 1 \geq \frac{9}{{(a+b+c)}^2}$

$\Leftrightarrow \frac{2}{3} \geq \frac{6}{{(a+b+c)}^2} \ldots (1)$

  • By $AM \geq GM$

$\frac{a^2+b^2}{2} \geq {ab} , \frac{b^2+c^2}{2} \geq {bc} , \frac{c^2+a^2}{2} \geq {ca}$

$\Leftrightarrow {a^2+b^2+c^2} \geq {ab+bc+ac}$

$\Leftrightarrow {a^2+b^2+c^2 + 2(ab+bc+ca)} \geq {3(ab+bc+ca)}$

$\Leftrightarrow {(a+b+c)}^2 \geq {3(ab+bc+ca)}$

$\Leftrightarrow \frac{1}{3} \geq \frac{ab+bc+ca}{{(a+b+c)}^2} \ldots (2)$

  • $(1) + (2)$

$1 \geq \frac{ab+bc+ca+6}{{(a+b+c)}^2} \ldots (3)$

  • By ${abc} \geq 1$

$\frac{ab+bc+ca+6}{{(a+b+c)}^2} \geq \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+6}{{(a+b+c)}^2} \ldots (4)$

  • By Cauchy-Schwarz

${(\frac{1}{a}+1+1)(a^3+b^2+c^2)} \geq {(a+b+c)}^2 \Leftrightarrow \frac{\frac{1}{a}+2}{(a+b+c)^2} \geq \frac{1}{a^3+b^2+c^2} \ldots (5)$

${(\frac{1}{b}+1+1)(a^2+b^3+c^2)} \geq {(a+b+c)}^2 \Leftrightarrow \frac{\frac{1}{b}+2}{(a+b+c)^2} \geq \frac{1}{a^2+b^3+c^2} \ldots (6)$

${(\frac{1}{c}+1+1)(a^2+b^2+c^3)} \geq {(a+b+c)}^2 \Leftrightarrow \frac{\frac{1}{c}+2}{(a+b+c)^2} \geq \frac{1}{a^2+b^2+c^3} \ldots (7)$

  • $(5) + (6) + (7)$

$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+6}{{(a+b+c)}^2} \geq \frac{1}{a^3+b^2+c^2}+\frac{1}{a^2+b^3+c^2}+\frac{1}{a^2+b^2+c^3}\ldots (8)$

  • $Q.E.D.$ by $(3),(4),(8)$

 

CMIIW :)

Edited by minohcity
  • Upvote 3

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