Jump to content
Sign in to follow this  
-_-

[JBMO 2015 No 2] $a+b+c=3$, nilai minimum dari...

Recommended Posts

$a,b,c$ bilangan riil positif sehingga $a+b+c=3$. Tentukan nilai minimum dari $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$


Share this post


Link to post
Share on other sites

tolong koreksi bila ada yg salah


\begin{align}
& \frac{2}{a}+\frac{2}{b}+\frac{2}{c}-a^2-b^2-c^2 \text{, AM-HM} \\ \\ 
& \geq \frac{9}{\frac{a+b+c}{2}}-(a^2+b^2+c^2) \\ \\
& = 6-((a+b+c)^2-2(ab+bc+ac)) \\ \\
& = 6-9+2(ab+bc+ac)  \text{, AM-HM}\\ \\ 
& \geq -3+2\left(\frac{9}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}}\right) \\ \\
& = -3+2\left(\frac{9}{\frac{a+b+c}{abc}}\right) \\ \\
& = -3+6abc \\ \\ 
& = -3(1-2abc) \text{, disini haruslah $(abc)$ maksimum}\\ \\
\text{perhatikan bahwa :} \frac{a+b+c}{3} \geq \sqrt[3]{abc} \implies 1 \geq abc \text{, maka} \\ \\
& \geq -3(1-2) \\ \\
& = 3
\end{align}
jadi nilai minimum dari $\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$ adalah $3$

Share this post


Link to post
Share on other sites

tolong koreksi bila ada yg salah

\begin{align}

& \frac{2}{a}+\frac{2}{b}+\frac{2}{c}-a^2-b^2-c^2 \text{, AM-HM} \\ \\ 

& \geq \frac{9}{\frac{a+b+c}{2}}-(a^2+b^2+c^2) \\ \\

& = 6-((a+b+c)^2-2(ab+bc+ac)) \\ \\

& = 6-9+2(ab+bc+ac)  \text{, AM-HM}\\ \\ 

& \geq -3+2\left(\frac{9}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}}\right) \\ \\

& = -3+2\left(\frac{9}{\frac{a+b+c}{abc}}\right) \\ \\

& = -3+6abc \\ \\ 

& = -3(1-2abc) \text{, disini haruslah $(abc)$ maksimum}\\ \\

\text{perhatikan bahwa :} \frac{a+b+c}{3} \geq \sqrt[3]{abc} \implies 1 \geq abc \text{, maka} \\ \\

& \geq -3(1-2) \\ \\

& = 3

\end{align}

jadi nilai minimum dari $\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$ adalah $3$

ineq terakhir tandanya kebalik, harusnya $-3+6abc\leq -3+6(1)=3$ ._.

Share this post


Link to post
Share on other sites

 

tolong koreksi bila ada yg salah

\begin{align}

& \frac{2}{a}+\frac{2}{b}+\frac{2}{c}-a^2-b^2-c^2 \text{, AM-HM} \\ \\ 

& \geq \frac{9}{\frac{a+b+c}{2}}-(a^2+b^2+c^2) \\ \\

& = 6-((a+b+c)^2-2(ab+bc+ac)) \\ \\

& = 6-9+2(ab+bc+ac)  \text{, AM-HM}\\ \\ 

& \geq -3+2\left(\frac{9}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}}\right) \\ \\

& = -3+2\left(\frac{9}{\frac{a+b+c}{abc}}\right) \\ \\

& = -3+6abc \\ \\ 

& = -3(1-2abc) \text{, disini haruslah $(abc)$ maksimum}\\ \\

\text{perhatikan bahwa :} \frac{a+b+c}{3} \geq \sqrt[3]{abc} \implies 1 \geq abc \text{, maka} \\ \\

& \geq -3(1-2) \\ \\

& = 3

\end{align}

jadi nilai minimum dari $\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$ adalah $3$

ineq terakhir tandanya kebalik, harusnya $-3+6abc\leq -3+6(1)=3$ ._.

 

haduh pantesan aja :v ya sudahlah ...

Share this post


Link to post
Share on other sites



Misalkan $X=a^{2}+b^{2}+c^{2}$ dan $Y=ab+bc+ca$. Dengan ketaksamaan AM-GM

$$Y^{2}=(ab+bc+ca)^{2}\geq 3(ab\cdot bc+bc\cdot ca+ca\cdot ab)=3abc(a+b+c)=9abc$$


dan


$$81=(a+b+c)^{4}=(X+2Y)^{2}=\left( \frac{2X+Y+3Y}{2}\right) ^{2}\geq (2X+Y)\cdot 3Y.$$


Menggunakan dua ketaksamaan di atas diperoleh


$$\begin{eqnarray*} 3+X &=&\frac{(a+b+c)^{2}+3X}{3}=\frac{2Y+4X}{3}=\frac{2}{3}\left( 2X+Y\right) \\ &\leq &\frac{2}{3}\cdot \frac{81}{3Y}=\frac{18}{Y}\leq \frac{2Y}{abc}. \end{eqnarray*}$$


Ketaksamaan terakhir ekivalen dengan


$$\begin{eqnarray*}3 &\leq &\frac{2Y}{abc}-X=2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-(a^{2}+b^{2}+c^{2}) \\ &=&\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}. \end{eqnarray*}$$


Kesamaan terjadi jika dan hanya jika $a=b=c=1$.


  • Upvote 1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this  

×