Točno
6. kolovoza 2012. 20:57 (11 godine, 11 mjeseci)
For each integer
![n\geqslant2](/media/m/7/d/e/7de025e723264e181387bc1b65daa65b.png)
, determine the largest real constant
![C_n](/media/m/c/4/7/c4741f92cb04e74aca4942700ae6f7bf.png)
such that for all positive real numbers
![a_1, \ldots, a_n](/media/m/4/6/2/4620bab4413c05b6365d3f7e207d102a.png)
we have
%V0
For each integer $n\geqslant2$, determine the largest real constant $C_n$ such that for all positive real numbers $a_1, \ldots, a_n$ we have $$\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+C_n\cdot(a_1-a_n)^2\mbox{.}$$
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Uvrstimo
![a_1=2](/media/m/3/6/6/3669830e001093aa3c8a75a8aedfaff3.png)
,
![a_2=...=a_{n-1}=1](/media/m/0/d/7/0d71ec6e146e7a9a1432ef805ca33654.png)
i
![a_n=0](/media/m/6/e/0/6e094b8b7f6f3b2092516840931cc716.png)
pa vidimo da je
![C_n \leq \frac{1}{2n}](/media/m/0/1/4/01497c59dd5e3ba612f531796d0b149b.png)
. Dokažimo nejednakost za
![C_n = \frac{1}{2n}](/media/m/f/e/f/fefb6c4488712532e10d112855fa16aa.png)
.
Neka je
![\frac{a_1+...+a_n}{n} =a](/media/m/1/7/7/1775a7aa5bfc9224972605a9c69b8b3c.png)
i
![x_i=a_i-a](/media/m/d/0/e/d0ef9ab3398c830265a271e42ccd63b2.png)
. Tada je očito
![x_1+...+x_n=0](/media/m/6/c/d/6cd74e947b85f6abb81e58a44ad8384b.png)
, a nejednakost je ekvivalentna s:
![\frac{1}{n} \sum x_i^2 \ge \frac{1}{2n}(x_1-x_n)^2](/media/m/e/a/2/ea23bc54e78d76a1e15750afa9ae35f2.png)
![\Leftrightarrow 2\sum x_i^2 \ge (x_1-x_n)^2](/media/m/c/f/5/cf53c83a38d6c99055a8835e5926c037.png)
Što slijedi iz:
%V0
Uvrstimo $a_1=2$, $a_2=...=a_{n-1}=1$ i $a_n=0$ pa vidimo da je $C_n \leq \frac{1}{2n}$. Dokažimo nejednakost za $C_n = \frac{1}{2n}$.
Neka je $\frac{a_1+...+a_n}{n} =a$ i $x_i=a_i-a$. Tada je očito $x_1+...+x_n=0$, a nejednakost je ekvivalentna s:
$\frac{1}{n} \sum x_i^2 \ge \frac{1}{2n}(x_1-x_n)^2$
$\Leftrightarrow 2\sum x_i^2 \ge (x_1-x_n)^2$
Što slijedi iz:
$2\sum x_i^2 \ge 2(x_1^2+x_n^2)=(x_1+x_n)^2+(x_1-x_n)^2 \ge (x_1-x_n)^2$
16. prosinca 2012. 21:27 | grga | Točno |