Also was in KSMO(existed but not 3-variable inequality, but more general form) 3 Variable inequality problems seem to be minor to examiner as it has been investigated too much(uvw pqr sos mv and...) The level of KJMO would apparently have been decreased because of the policy of junior problems that "Focus more on the middle school curriculum" So that preliminary even had a lot of perfect scorers

Skravin wrote: Also was in KSMO(existed but not 3-variable inequality, but more general form) 3 Variable inequality problems seem to be minor to examiner as it has been investigated too much(uvw pqr sos mv and...) what was general form?

khyeon wrote: Skravin wrote: Also was in KSMO(existed but not 3-variable inequality, but more general form) 3 Variable inequality problems seem to be minor to examiner as it has been investigated too much(uvw pqr sos mv and...) what was general form? To be brief the form was similar to greedy algorithm that: Prove that for positive real numbers $a_1, a_2, \cdot \cdot \cdot a_n$, satisfying $a_1 \geq a_2 \geq \cdot \cdot \cdot a_n$ and for arbitrary positive real numbers $b_1, b_2, \cdot \cdot \cdot b_n$, $\frac{a_1 b_1+a_2 b_2+\cdot \cdot \cdot +a_n b_n}{a_1+a_2+\cdot \cdot \cdot +a_n} \leq max(\frac{b_1}{1}, \frac{b_1+b_2}{2}, \cdot \cdot \cdot \frac{b_1+b_2+\cdot \cdot \cdot+b_n}{n})$ I proved some with mathematical induction but not thought to be that brief so wouldn't post my solution... also, the problems used more geometry(3/8), which is my weak point(seems to be affected by the IMO style that tends to make more geometry problems)

Skravin wrote: Also was in KSMO(existed but not 3-variable inequality, but more general form) 3 Variable inequality problems seem to be minor to examiner as it has been investigated too much(uvw pqr sos mv and...) The level of KJMO would apparently have been decreased because of the policy of junior problems that "Focus more on the middle school curriculum" So that preliminary even had a lot of perfect scorers yes, at first round test.. but i can't expect cutline of second test I solved 5 problems, but I'm expecting bronze or lower silver It was easiest, even easier than 2014

khyeon wrote: Skravin wrote: Also was in KSMO(existed but not 3-variable inequality, but more general form) 3 Variable inequality problems seem to be minor to examiner as it has been investigated too much(uvw pqr sos mv and...) The level of KJMO would apparently have been decreased because of the policy of junior problems that "Focus more on the middle school curriculum" So that preliminary even had a lot of perfect scorers yes, at first round test.. but i can't expect cutline of second test I solved 5 problems, but I'm expecting bronze or lower silver It was easiest, even easier than 2014 junior 2nd has really high cutline even in 2012(when the KMO was the hardest ever), cutline for gold was 5...

khyeon wrote: positive integers $a_1, a_2, . . . , a_9$ satisfying $a_1+a_2+ . . . +a_9 =90$ find maximum of $$\frac{1^{a_1} \cdot 2^{a_2} \cdot . . . \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot . . . \cdot a_9!}$$

btw the solution would based on $a_1+a_2+\cdot \cdot \cdot +a_n=n(n+1)$ and investigate I assume? 1.one thing is apparent:$a_1=1$($a_1 \geq 2$ wouldn't give any advance to formula.)

khyeon wrote: positive integers $a_1, a_2, . . . , a_9$ satisfying $a_1+a_2+ . . . +a_9 =90$ find maximum of $$\frac{1^{a_1} \cdot 2^{a_2} \cdot . . . \cdot 9^{a_9}}{a_1! \cdot a_2! \cdot . . . \cdot a_9!}$$

Given positive integers \(a_1, a_2, \ldots, a_9\) satisfying \(a_1 + a_2 + \cdots + a_9 = 90\), we aim to maximize the expression \[ \frac{1^{a_1} 2^{a_2} \cdots 9^{a_9}}{a_1! a_2! \cdots a_9!}. \] Since there are only possibly many cases for $a_1$, $\ldots$, $a_9$ with $a_1 + \cdots + a_9 = 90$, there must exist maximum for the expression under consideration. Hence, assume that the expression \(\frac{1^{a_1} 2^{a_2} \cdots 9^{a_9}}{a_1! a_2! \cdots a_9!}\) achieved its maximum. For this, consider any \(1 \le i, j \le 9\). Letting \(x = a_i\) and \(y = a_j\), we have the following inequalities: \[ \frac{i^{x-1} j^{y+1}}{(x-1)!(y+1)!} \le \frac{i^x j^y}{x!y!}, \quad \frac{i^{x+1} j^{y-1}}{(x+1)!(y-1)!} \le \frac{i^x j^y}{x! y!}. \]Namely, \[ \frac{a_i}{i} \le \frac{a_j}{j} + \frac{1}{j}, \quad \frac{a_j}{j} \le \frac{a_i}{i} + \frac{1}{i}. \]This is obtained when \(\frac{a_i}{i} = c\). Given the constraint, we find that \(c = 2\). Suppose there is another case for the inequalities. Then at least one \(\frac{a_i}{i}\) should be less than or equal to \(2 - \frac{1}{i}\). At the same time, at least one \(\frac{a_j}{j}\) should be no less than \(2 + \frac{1}{j}\). Therefore, \(2 + \frac{1}{j} \le 2 - \frac{1}{i} + \frac{1}{i} = 2\), which is a contradiction. Hence, \(\frac{a_i}{i} = 2\) is the unique solution for the inequalities. It follows that the maximum is \[ \frac{1^2 2^4 3^6 \cdots 9^{18}}{2! 4! \cdots 18!}. \]