Let $p$, $q$ be two integers, $q\geq p\geq 0$. Let $n \geq 2$ be an integer and $a_0=0, a_1 \geq 0, a_2, \ldots, a_{n-1},a_n = 1$ be real numbers such that \[ a_{k} \leq \frac{ a_{k-1} + a_{k+1} } 2 , \ \forall \ k=1,2,\ldots, n-1 . \] Prove that \[ (p+1) \sum_{k=1}^{n-1} a_k^p \geq (q+1) \sum_{k=1}^{n-1} a_k^q . \]
Problem
Source: Romanian IMO TST 2006, day 4, problem 4
Tags: inequalities, function, integration, calculus, derivative, inequalities unsolved
xirti
21.05.2006 14:49
Remark: It is equivalent with : Given a convex function $f: [0,1]\to \mathbb{R}$ with $f(0)=0, f(1)=1$, prove that $(p+1)\int_0^1f^p(x)dx\le (q+1)\int_0^1f^q(x)dx$. Can someone prove the ineq for $a_k=\frac k n$ and $q=p+1$? In contest I reduced the problem to this case but I didn't have enough time to finish it. If the answer is positive I can write my (piece of ) solution.
zhaobin
22.05.2006 07:20
xirti wrote: Remark: It is equivalent with : Given a convex function $f: [0,1]\to \mathbb{R}$ with $f(0)=0, f(1)=1$, prove that $(p+1)\int_0^1f^p(x)dx\le (q+1)\int_0^1f^q(x)dx$. Can someone prove the ineq for $a_k=\frac k n$ and $q=p+1$? In contest I reduced the problem to this case but I didn't have enough time to finish it. If the answer is positive I can write my (piece of ) solution. Nice idea. what do you mean by "If the answer is positive "? means $f(x)>0$ PS:can you post you solution?
xirti
29.05.2006 19:41
xirti wrote: Can someone prove the ineq for $a_k=\frac k n$ and $q=p+1$? If the answer of this question is positive... (positive means affirmative; yes)
hxy09
06.07.2009 12:07
Anybody interests?I read the official solution of this problem,I think it was amazing.And this is why I bounced the topic to the top to see your nice solutions
gb2124
25.08.2009 07:26
Where do you find the official solutions,Can you give a link?
hxy09
30.08.2009 07:48
gb2124 wrote: Where do you find the official solutions,Can you give a link? It's for you
Attachments:
Albanian Eagle
30.08.2009 11:14
It's actually quite simple in the integral version mentioned above. The inequality we have to prove is \[ (p+1)\int_0^1 f^p(x)dx\geq (q+1)\int_0^1f^q(x)dx\] Where $ f$is convex and increasing. I will assume that $ f$ is differentiable even though it can be avoided The inequality is equivalent to \[ \int_0^1x^qdx\int_0^1f^p(x)dx\geq \int_0^1x^pdx\int_0^1f^q(x)dx\] and in turn equivalent to \[ \int_0^1\int_0^1 f^q(x)f^p(y)f'(x)dxdy\geq \int_0^1\int_0^1 f^q(x)f^p(y)f'(y)dxdy\] which can be rewritten as \[ \int_0^1\int_0^1 f^p(x)f^p(y)(f^{q-p}(x)-f^{q-p}(y))(f'(x)-f'(y))dxdy\geq 0\] Because $ f$ is convex $ f'$ is increasing and we are done.
gb2124
30.08.2009 12:18
Dear Albanian eagle: I don't quite understand the solution(since i'm not good at integral). Since ak is non continuable,how can you use that? And furthermore i don't understand the last sentence you wrote. Why is that inequality is true since f is convex and f' is increases. Can you explain more to me? Thanks in advance. And to hxy09 Thanks for the material.
Albanian Eagle
30.08.2009 12:52
dear gb2124, I proved the inequality claimed in the second post by xirti. He claimed an equivalence to the original problem. From my solution you can get the original problem if you convert each step to it's discrete analog. Keep in mind that integrals are just "continuous summations". Derivatives will correspond to finite differences etc. If you do that you'll see a motivation as to why one uses the Abel summation in the official solution. As to the last line if you look carefully the function being integrated is non-negative and therefore the integral is non-negative.
gb2124
30.08.2009 13:30
Dear albanian eagle Oh yes! I forgot the condition that f is also increasing too. I think you are right.