Točno
15. ožujka 2016. 23:45 (8 godine, 4 mjeseci)
Let
![w, x, y, z](/media/m/e/4/6/e46dd38573fcbfad752112ec556e0b86.png)
are non-negative reals such that
![wx + xy + yz + zw = 1](/media/m/2/1/f/21f770506292fade7511ed234a9fbb8c.png)
. Show that
![\frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}](/media/m/4/c/8/4c873f953fca8e23e174899fa5090807.png)
.
%V0
Let $w, x, y, z$ are non-negative reals such that $wx + xy + yz + zw = 1$. Show that
$$\frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$$.
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Uocimo da uvijet mozemo zapisati kao
![1=x(w+y)+z(w+y)=(x+z)(w+y)](/media/m/3/5/f/35f87563ddd3feeae0dbf314554b3d48.png)
Prema Holderovoj nejednakosti vrijedi:
![\displaystyle \sum\limits_{cyc}{ \frac{w^3}{x+y+z}} \cdot 3\sum\limits_{cyc}{x} \cdot (1+1+1+1) \geqslant (x+y+z+w)^3](/media/m/5/1/7/517052592615dfa8d97f5008fc079ae2.png)
pa nakon djeljenja prethodnog izraza sa
![12(x+y+z+w)](/media/m/a/3/a/a3a6bf702fd7b6071430b283f3cf94cd.png)
dobivamo
![\displaystyle \sum\limits_{cyc}{ \frac{w^3}{x+y+z}} \geqslant \frac{(x+y+z+w)^2}{12}](/media/m/1/7/0/170a3afc8d6e799a0c24246d01e6110e.png)
pa je dovoljno pokazati
![\displaystyle \frac{(x+y+z+w)^2}{12} \geqslant \frac{1}{3}](/media/m/c/5/3/c53a48859204f2a099578a6801b91841.png)
odnosno
![\displaystyle (x+y+z+w)^2 \geqslant 4](/media/m/b/f/b/bfb74cb3e61ecffc3e635a7a2580e9e3.png)
nakon korjenovanja, sto smijemo jer su
![x,y,z,w](/media/m/e/4/e/e4ea20c9f79f577f9eb540fe77d9d317.png)
pozitivni dobivamo
![\displaystyle (x+y+z+w) \geqslant 2](/media/m/a/d/d/addcc17f1f47bacbe2c9ab40a24f11a8.png)
oznacimo sada
![x+z=a](/media/m/5/5/a/55af34134e9555d6eb12f1a6ef8050e9.png)
i
![y+w=b](/media/m/3/4/6/34679863d8b0fd5b71fbf0a33ce9d589.png)
tada vrijedi
![ab=1](/media/m/9/3/b/93b5aa9bf76906d64840d051175ff759.png)
pa jednakost prelazi u
![\displaystyle (a+b) \geqslant 2](/media/m/a/0/9/a0950a414b14764d7b7835ce0c7a26bb.png)
sto vrijedi po A-G nejdnakosti
%V0
Uocimo da uvijet mozemo zapisati kao $ 1=x(w+y)+z(w+y)=(x+z)(w+y) $
Prema Holderovoj nejednakosti vrijedi:
$ \displaystyle \sum\limits_{cyc}{ \frac{w^3}{x+y+z}} \cdot 3\sum\limits_{cyc}{x} \cdot (1+1+1+1) \geqslant (x+y+z+w)^3 $
pa nakon djeljenja prethodnog izraza sa $ 12(x+y+z+w) $ dobivamo
$ \displaystyle \sum\limits_{cyc}{ \frac{w^3}{x+y+z}} \geqslant \frac{(x+y+z+w)^2}{12} $
pa je dovoljno pokazati
$ \displaystyle \frac{(x+y+z+w)^2}{12} \geqslant \frac{1}{3} $
odnosno
$ \displaystyle (x+y+z+w)^2 \geqslant 4 $
nakon korjenovanja, sto smijemo jer su $ x,y,z,w $ pozitivni dobivamo
$ \displaystyle (x+y+z+w) \geqslant 2 $
oznacimo sada $ x+z=a $ i $ y+w=b $ tada vrijedi $ ab=1 $ pa jednakost prelazi u
$ \displaystyle (a+b) \geqslant 2 $ sto vrijedi po A-G nejdnakosti
1. travnja 2016. 17:16 | grga | Točno |