Točno
22. travnja 2012. 18:04 (12 godine, 1 mjesec)
Za pozitivne brojeve
![a_1](/media/m/6/1/7/6173ac27c63013385bea9def9ff2b61e.png)
,
![a_2](/media/m/4/0/1/401f4cdfec59fba73ae32fa6769c72cb.png)
,
![\dots](/media/m/3/6/1/36118a223c1f6e75548277354fbabc8a.png)
,
![a_n](/media/m/1/f/f/1ff6f81c68b9c6fb726845c9ce762d7a.png)
,
![n \geq 2](/media/m/2/1/f/21fe2458de6d1580c44fd06e0fac11bb.png)
označimo
![a_1 + a_2 + \dots + a_n = s](/media/m/8/1/7/8177c2e7f3170ceca938aa7fd1d4c537.png)
. Dokažite nejednakost
%V0
Za pozitivne brojeve $a_1$, $a_2$, $\dots$, $a_n$, $n \geq 2$ označimo $a_1 + a_2 + \dots + a_n = s$. Dokažite nejednakost
$$\dfrac{a_1}{s - a_1} + \dfrac{a_2}{s - a_2} + \dots + \dfrac{a_n}{s - a_n} \geq \dfrac{n}{n - 1}\text{.}$$
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Definirajmo nizove
![A = \left\{ a_1, a_2, a_3, \cdots, a_n \right\}](/media/m/4/c/8/4c80ac6b9ba23cb791699ecd30550f0a.png)
i
![B = \left\{\dfrac{1}{s-a_1}, \dfrac{1}{s-a_2}, \cdots, \dfrac{1}{s-a_n}\right\}](/media/m/7/4/d/74d4967c07e35c69fcc4b4a478a1b4e7.png)
BSOMP
![a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n](/media/m/d/4/b/d4b8ba207358bb56eb7f806ba20d04d0.png)
. Onda vrijedi i
![\dfrac{1}{s-a_1} \geq \dfrac{1}{s-a_2} \geq \cdots \geq \dfrac{1}{s-a_n}](/media/m/f/9/c/f9cd05a1992c2f936392f9326a0a3e3a.png)
, pa iz MPV nejednakosti slijedi da je
![LHS \geqslant P](/media/m/d/e/8/de876f6e5c5d119c6450cf11d5ab1986.png)
, gdje je P bilo koja premutacija umnožaka
![A_i*B_i](/media/m/7/d/9/7d9eec094de65b1e8e31bd0097091921.png)
.
Sada vrijedi:
![(n-1)*LHS \geqslant \frac{a_2+a_3+\cdots+a_n}{s-a_1} + \frac{a_1 + a_3 + a_4 + \cdots + a_n}{s-a_2} + \cdots + \frac{a_1+a_2+\cdots+a_{n-1}}{s-a_n}](/media/m/9/0/b/90bf44116f317fca404ce33fa248fd8f.png)
![= 1 + 1 + 1 + \cdots + 1 = n](/media/m/d/0/2/d0286cb12cfcd7f38820b9a80a05d5d3.png)
%V0
Definirajmo nizove $A = \left\{ a_1, a_2, a_3, \cdots, a_n \right\}$ i $B = \left\{\dfrac{1}{s-a_1}, \dfrac{1}{s-a_2}, \cdots, \dfrac{1}{s-a_n}\right\}$
BSOMP $a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n$. Onda vrijedi i $\dfrac{1}{s-a_1} \geq \dfrac{1}{s-a_2} \geq \cdots \geq \dfrac{1}{s-a_n}$, pa iz MPV nejednakosti slijedi da je $ LHS \geqslant P $, gdje je P bilo koja premutacija umnožaka $A_i*B_i$.
Sada vrijedi:
$$(n-1)*LHS \geqslant \frac{a_2+a_3+\cdots+a_n}{s-a_1} + \frac{a_1 + a_3 + a_4 + \cdots + a_n}{s-a_2} + \cdots + \frac{a_1+a_2+\cdots+a_{n-1}}{s-a_n}$$
$$ = 1 + 1 + 1 + \cdots + 1 = n $$
$$\Leftrightarrow LHS \geqslant \frac{n}{n-1}\text{ }\blacksquare$$
6. svibnja 2012. 22:34 | ikicic | Točno |
22. travnja 2012. 19:04 | grga | Točno |