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timo123

ketaksamaan $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

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Misalkan $x,y,z$ bilangan real positif. Diberikan bilangan real $k\geq 1$. Buktikan atau buktikan salah $$\begin{align*}\frac{1}{k+2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+&\frac{1}{kx+y+z}+\frac{1}{x+ky+z}+\frac{1}{x+y+kz}\geq \\&\frac{1}{(k+1)x+y}+\frac{1}{(k+1)x+z}+\frac{1}{(k+1)y+x}+\frac{1}{(k+1)y+z}+\frac{1}{(k+1)z+x}+\frac{1}{(k+1)z+y}\end{align*}$$


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Ini sebenarnya cuma Schur biasa, dikasih integral..


 


Dari $\sum a^k(a-b)(a-c) \ge 0$, substitusikan $a=t^x$ dimana $t\ge 0$, lalu bagi dengan $t$, trus integral dalam variabel $t$ rentang $0$ sampe $1$.


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Ini sebenarnya cuma Schur biasa, dikasih integral..

 

Dari $\sum a^k(a-b)(a-c) \ge 0$, substitusikan $a=t^x$ dimana $t\ge 0$, lalu bagi dengan $t$, trus integral dalam variabel $t$ rentang $0$ sampe $1$.

 

mau dong solusi lengkapnya pak, udah setaun ga bisa2 ini..

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Oke, dari Schur $$\sum_{cyc} a^k(a-b)(a-c) \ge 0 $$dan substitusi $a=t^x,b=t^y,c=t^z$, kita dapat:


 


$$\sum t^{(k+2)x} + t^{kx+y+z} \ge \sum t^{(k+1)x+y} + t^{(k+1)x+z}$$


 


Bagi kedua ruas dengan $t$, kita dapat


 


$$\sum t^{(k+2)x-1} + t^{kx+y+z-1} \ge \sum t^{(k+1)x+y-1} + t^{(k+1)x+z-1}$$


 


Kan kalau $f(x) \ge g(x) \ge 0$ maka kalo kita integralin masih valid kan ya ketaksamaannya.


 


Jadi kita punya:


 


$$\int_0^1 \sum \left( t^{(k+2)x-1} + t^{kx+y+z-1}\right) dt \ge \int_0^1 \sum \left(t^{(k+1)x+y-1} + t^{(k+1)x+z-1}\right) dt$$


 


$$\sum \frac{1}{(k+2)x} t^{(k+2)x}|_0^1 + \frac{1}{kx+y+z} t^{kx+y+z}|_0^1 \ge \sum \frac{1}{(k+1)x+y} t^{(k+1)x+y}|_0^1 + \frac{1}{(k+1)x+z} t^{(k+1)x+z}|_0^1$$


 


yang udah sama persis sama soalnya


 


PS : notasinya mungkin rada ngawur2 gitu, mungkin susah dipahami cyc2nya gitu, tapi kalo ditulis panjang harusnya paham kok. Males soalnya nulis panjang..


Edited by blajaran

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Ubah ke bentuk Vornicu-Schur juga bisa: $$ \sum \left(\frac{1}{(k+2)x} + \frac{1}{kx+y+z} \right) \frac{(x-y)(x-z)}{((k+1)x + y)((k+1)x + z)} \geq 0.$$


 


Bukti bahwa bisa diubah: :p


 


Misalkan $a = (k+2)x$, $b = kx + y + z$, $c = (k+1)x + y$ dan $d = (k+1)x + z$, maka $a+b=c+d$ dan $cd-ab = (x-y)(x-z)$. Lalu $$\frac 1 a + \frac 1 b - \frac 1 c - \frac 1 d = \frac{(a+b)(x-y)(x-z)}{abcd}.$$


Edited by candhakkeplekkegebuk

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