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Ini ketaksamaan dua variabel kok

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Untuk semua $a,b$ bilangan real positif, buktikan $$\frac{1}{4a}+\frac{3}{a+b}+\frac{1}{4b}\geq \frac{4}{3a+b}+\frac{4}{3b+a}$$

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Karena ineqnya homogen, wlog $\[a+b=1]\$
Maka, $\[(a+b)^2 \geq 4ab -> 1 \geq 4ab\]$

 

Dari wlog kita maka ineqnya jadi

 

$\[3+\frac{1}{4a}+\frac{1}{4b} \geq \frac{4}{2a+1}+\frac{4}{2b+1}\]$

$\[3+\frac{a+b}{4ab} \geq \frac{8a+4+8b+4}{(2a+1)(2b+1)}\]$

$\[3+\frac{1}{4ab} \geq \frac{16}{4ab+2(a+b)+1}\]$

$\[3+\frac{1}{4ab} \geq \frac{16}{4ab+3}\]$

 

Misal $\[k=4ab\]$, maka $\[k \leq 1\]$ atau $\[k-1 \leq 0\]$

 

Lihat bahwa 

$\[3+\frac{1}{4ab} \geq \frac{16}{4ab+3}\]$

$\[3+\frac{1}{k} \geq \frac{16}{x+3}\]$

$\[\frac{k+3}{k} \geq \frac{16}{x+3}\]$

 

Kali silang, kita dapat

$\[(k+3)^2 \geq 16k\]$

$\[k^2+6k+9 \geq 16k\]$
$\[k^2-10k+9 \geq 0\]$

$\[(k-9)(k-1) \geq 0\]$

 

Karena $\[k-1 \leq 0\]$, maka $\[k-9 < 0\]$, artinya jelas ineq terakhir kita benar

kesamaan terjadi bila $\[k-1=0 -> k=1\]$ yaitu ketika $\[a=b=\frac{1}{2}\]$ QED

 

 

 

 

 

 

Edited by DickJessen
LATEX gimana caranya
  • Upvote 1

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@DickJessen: kalau sudah pakai syntax dollar tidak perlu memakai syntax \[ lagi ._.

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