I tried to search SRMC problems,but i didn't find them(I found only SRMC 2006).Maybe someone know where on this site i could find SRMC problems?I have all SRMC problems,if someone want i could post them, Here is one of them,this is one nice inequality from first SRMC: Let $ n$ be an integer with $ n>2$ and $ a_{1},a_{2},\dots,a_{n}\in R^{+}$.Given any positive integers $ t,k,p$ with $ 1<t<n$,set $ m=k+p$,prove the following inequalities: a) $ \frac{a_{1}^{p}}{a_{2}^{k}+a_{3}^{k}+\dots+a_{t}^{k}}+\frac{a_{2}^{p}}{a_{3}^{k}+a_{4}^{k}+\dots+a_{t+1}^{k}}+\dots+\frac{a_{n-1}^{p}}{a_{n}^{k}+a_{1}^{k}+\dots+a_{t-2}^{k}}+\frac{a_{n}^{p}}{a_{1}^{k}+a_{2}^{k}+\dots+a_{t-1}^{k}}\geq\frac{(a_{1}^{p}+a_{2}^{p}\dots+a_{n}^{p})^{2}}{(t-1) ( a_{1}^{m}+a_{2}^{m}+\dots+a_{n}^{m})}$ b)$ \frac{a_{2}^{k}+a_{3}^{k}\dots+a_{t}^{k}}{a_{1}^{p}}+\frac{a_{3}^{k}+a_{4}^{k}\dots+a_{t+1}^{k}}{a_{2}^{p}}+\dots+\frac{a_{1}^{k}+a_{2}^{k}\dots+a_{t-1}^{k}}{a_{n}^{p}}\geq\frac{(t-1)(a_{1}^{k}+a_{2}^{k}\dots+a_{n}^{k})^{2}}{( a_{1}^{m}+a_{2}^{m}+\dots+a_{n}^{m})}$
Problem
Source: From SRMC
Tags: inequalities, search, inequalities proposed, SRMC 2002
05.09.2007 23:17
Erken what is SRMC ? :
06.09.2007 15:27
galois01 wrote: Erken what is SRMC ? : It is Silk Road Mathematics Competiton .It is very nice olympiad and quite hard olympiad.
06.09.2007 17:47
Erken wrote: galois01 wrote: Erken what is SRMC ? : It is Silk Road Mathematics Competiton .It is very nice olympiad and quite hard olympiad. Thank's a lot for your explanation it's the first time I hear about this olympiad
16.09.2007 18:56
Nobody?
28.04.2019 04:54
$[a]$ For the first part, denote the left-hand-side by $S$. Using Cauchy-Schwarz inequality, we have, $$ S\left( \sum_{i=1}^p a_i^p(a_{i+1}^k+\dots+a_{i+t-1}^k)\right)\geq (a_1^p+\cdots+a_n^p)^2, $$where, the indices above are taken modulo $n$. In particular, the inequality boils down to proving, $$ (t-1)(a_1^m+\cdots+a_n^m)\geq (a_1^pa_2^k+a_1^pa_3^k+\cdots+a_1^pa_t^k+a_2^pa_3^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k). $$Observe that, both sides have, upon opening, $n(t-1)$ terms. Now, keeping in mind that, $m=k+p$, consider the vectors, $$ v_1=(\underbrace{a_1^k,a_1^k,\dots,a_1^k}_{n-1},\underbrace{a_2^k,\dots,a_2^k}_{n-1},\dots,\underbrace{a_n^k,\dots,a_n^k}_{n-1}), $$and, $$ v_2=(\underbrace{a_1^p,a_1^p,\dots,a_1^p}_{n-1},\underbrace{a_2^p,\dots,a_2^p}_{n-1},\dots,\underbrace{a_n^p,\dots,a_n^p}_{n-1}). $$The inequality above, is simply a result of rearrangement inequality, applied on $v_1$ and $v_2$. $[b]$ For the second part, break the each terms into sum of individual terms, that is, for instance, $$ \frac{{a_2^k + a_3^k + \cdots + a_t^k}}{{a_1^p}}=\frac{a_2^k}{a_1^p}+\frac{a_3^k}{a_1^p}+\cdots+\frac{a_t^k}{a_1^p}. $$Now, apply Cauchy-Schwarz on left-hand-side with weights $(a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)$, to get, $$ {\rm LHS}\cdot (a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)\geq ((t-1)(a_1^k+\cdots+a_n^k))^2. $$Now, using the fact that, $$ (a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)\leq (t-1)(a_1^m+\cdots+a_n^m), $$we arrive at the result.
16.08.2019 06:31
Erken wrote: galois01 wrote: Erken what is SRMC ? : It is Silk Road Mathematics Competiton .It is very nice olympiad and quite hard olympiad. Thanks. Are there any official websites?
16.08.2019 11:33
the aops contest collections are complete for SRMC Path: Contests \ International Contests \ Silk Road as a source for adding the missing problems in aops I used this page, I do not think that it is an official page but at least it has all the problems