Zadatak mozemo zapisati kao
mnozenjem sa
![\displaystyle (1+x)(1+y)(1+z)](/media/m/2/e/a/2ea8df6becd272817355bcdd714b2e60.png)
dobivamo ekvivalentnu nejednakost
![\displaystyle \sum\limits_{cyc}{x^3 \cdot (1+x)} \geqslant \frac{3}{4} \cdot (1+x)(1+y)(1+z)](/media/m/8/c/e/8ce6e3397c827833b29424256798174b.png)
zapisivati cu
![\displaystyle (a,b,c)](/media/m/5/7/4/57496bd3edc606abd4f7c7e610cc4b63.png)
sto mi oznacava
![\displaystyle \sum\limits_{sym}{ x^a y^b z^c}](/media/m/1/e/a/1ea4029e19d36a4f6f77ddba67325b84.png)
sada jednakost prelazi u
sada je nakon mnozenja sa
![8](/media/m/3/d/2/3d2c45264dbff498f9bcb16af5f83881.png)
dovoljno pokazati
![\displaystyle 4 (3,0,0) + 4 (4,0,0) \geqslant 3 (1,0,0)+ 3 (1,1,0)+2 (1,1,1)](/media/m/9/a/e/9ae679f5cae83667874c79ce284b7b21.png)
korištenjem
![xyz=1](/media/m/f/c/4/fc4d25ab80408fd281a61bf02f1c976d.png)
dobivamo sljedece nejednakosti koje vrijede po Muirheadu
![\displaystyle 3 (4,0,0) \geqslant 3 (2,1,1)=3(1,0,0)](/media/m/c/a/6/ca68339e10bbe32ed6ddf84a5f6b839d.png)
![\displaystyle (4,0,0) \geqslant ( \frac{5}{3} , \frac{5}{3} , \frac{2}{3} )=(1,1,0)](/media/m/3/c/d/3cd4bd20391c3c1bc0d4e4f80b30fe3b.png)
![\displaystyle 2(3,0,0) \geqslant 2 ( \frac{4}{3} , \frac{4}{3} , \frac{1}{3} )=2(1,1,0)](/media/m/0/8/2/082cd6f6d2b42fb50bc72c4f6143fddf.png)
![\displaystyle 2(3,0,0) \geqslant 2 (1,1,1)](/media/m/4/5/4/45421553bfa6ce9725e2b0079511c722.png)
Zbrajanjem slijedi nejednakost koju smo htjeli pokazati
%V0
Zadatak mozemo zapisati kao
$ \displaystyle \sum\limits_{cyc}{ \frac{x^3}{(1+y)(1+z)}} \geqslant \frac{3}{4} $
mnozenjem sa $ \displaystyle (1+x)(1+y)(1+z) $ dobivamo ekvivalentnu nejednakost
$ \displaystyle \sum\limits_{cyc}{x^3 \cdot (1+x)} \geqslant \frac{3}{4} \cdot (1+x)(1+y)(1+z) $
zapisivati cu $ \displaystyle (a,b,c) $ sto mi oznacava $ \displaystyle \sum\limits_{sym}{ x^a y^b z^c} $
sada jednakost prelazi u
$ \displaystyle \frac{1}{2} (3,0,0) + \frac{1}{2} (4,0,0) \geqslant \frac{3}{4} \cdot ( \frac{1}{2} (1,0,0)+ \frac{1}{2} (1,1,0)+ \frac{1}{3} (1,1,1) $
sada je nakon mnozenja sa $ 8 $ dovoljno pokazati
$ \displaystyle 4 (3,0,0) + 4 (4,0,0) \geqslant 3 (1,0,0)+ 3 (1,1,0)+2 (1,1,1) $
korištenjem $ xyz=1 $ dobivamo sljedece nejednakosti koje vrijede po Muirheadu
$ \displaystyle 3 (4,0,0) \geqslant 3 (2,1,1)=3(1,0,0) $
$ \displaystyle (4,0,0) \geqslant ( \frac{5}{3} , \frac{5}{3} , \frac{2}{3} )=(1,1,0) $
$ \displaystyle 2(3,0,0) \geqslant 2 ( \frac{4}{3} , \frac{4}{3} , \frac{1}{3} )=2(1,1,0) $
$ \displaystyle 2(3,0,0) \geqslant 2 (1,1,1) $
Zbrajanjem slijedi nejednakost koju smo htjeli pokazati