I solved it in the TST,but my solution was rather long,I am waiting for a short and nice solution here

Let me see if this works. EDIT: it doesn't, never mind.

official solution

sorry but I can not read the hand writing , can you use $ \text{\LaTeX}$ ? Thanks !

hxy09 wrote: official solution There's also another proof using induction. What is your proof at the TST, hxy09? Could you please post it here? Thanks.

can_hang2007 wrote: hxy09 wrote: official solution There's also another proof using induction. What is your proof at the TST, hxy09? Could you please post it here? Thanks. I used induction and derivative.But the proof is very long and not elegant. I will try to post it next week.

This is my ugly solution(the motivation is simple,but calculation is terrible)

Fang-jh wrote: Given an integer $ n\ge 2$, find the maximal constant $ \lambda (n)$ having the following property: if a sequence of real numbers $ a_{0},a_{1},a_{2},\cdots,a_{n}$ satisfies $ 0 = a_{0}\le a_{1}\le a_{2}\le \cdots\le a_{n},$ and $ a_{i}\ge\frac {1}{2}(a_{i + 1} + a_{i - 1}),i = 1,2,\cdots,n - 1,$ then $ (\sum_{i = 1}^n{ia_{i}})^2\ge \lambda (n)\sum_{i = 1}^n{a_{i}^2}.$ See my proof at: http://canhang2007.wordpress.com/2009/11/06/inequality-51-unknown-author/

can_hang2007 wrote: Fang-jh wrote: Given an integer $ n\ge 2$, find the maximal constant $ \lambda (n)$ having the following property: if a sequence of real numbers $ a_{0},a_{1},a_{2},\cdots,a_{n}$ satisfies $ 0 = a_{0}\le a_{1}\le a_{2}\le \cdots\le a_{n},$ and $ a_{i}\ge\frac {1}{2}(a_{i + 1} + a_{i - 1}),i = 1,2,\cdots,n - 1,$ then $ (\sum_{i = 1}^n{ia_{i}})^2\ge \lambda (n)\sum_{i = 1}^n{a_{i}^2}.$ See my proof at: http://canhang2007.wordpress.com/2009/11/06/inequality-51-unknown-author/ Hi can_hang2007, I am very sorry to tell you that wordpress is not available in China( at least in Hubei) Could you please paste your solution here,thank you so much

hxy09 wrote: can_hang2007 wrote: Fang-jh wrote: Given an integer $ n\ge 2$, find the maximal constant $ \lambda (n)$ having the following property: if a sequence of real numbers $ a_{0},a_{1},a_{2},\cdots,a_{n}$ satisfies $ 0 = a_{0}\le a_{1}\le a_{2}\le \cdots\le a_{n},$ and $ a_{i}\ge\frac {1}{2}(a_{i + 1} + a_{i - 1}),i = 1,2,\cdots,n - 1,$ then $ (\sum_{i = 1}^n{ia_{i}})^2\ge \lambda (n)\sum_{i = 1}^n{a_{i}^2}.$ See my proof at: http://canhang2007.wordpress.com/2009/11/06/inequality-51-unknown-author/ Hi can_hang2007, I am very sorry to tell you that wordpress is not available in China( at least in Hubei) Could you please paste your solution here,thank you so much I think you can use the site http://zend2.com to open wordpress blog. Now, I copy my proof here: We choose $ \displaystyle (a_{0},a_{1},a_{2},\ldots ,a_{n}) = (0,1,\ldots ,1).$ This sequence satisfies the given hypothesis and after substituting into the desired inequality, we get $ \displaystyle \lambda (n)\leq \dfrac{n(n + 1)^{2}}{4}.$ We will show that the maximal constant $ \displaystyle \lambda (n)$ is $ \displaystyle \dfrac{n(n + 1)^{2}}{4};$ that is, to show that the following inequality holds $ \displaystyle \left( \sum_{i = 1}^{n}ia_{i}\right) ^{2}\geq \dfrac{n(n + 1)^{2}}{4}\sum_{i = 1}^{n}a_{i}^{2}.$ We will prove this inequality by induction on $ \displaystyle n.$ For $ \displaystyle n = 2,$ we have $ \displaystyle (a_{1} + 2a_{2})^{2} - \frac {9}{2}(a_{1}^{2} + a_{2}^{2}) = \frac {1}{2}(a_{2} - a_{1})(7a_{1} - a_{2})\geq 0,$ because $ \displaystyle 2a_{1}\geq a_{0} + a_{2} = a_{2}.$ So, the inequality is true for $ \displaystyle n = 2.$ Now, assume that it holds for $ \displaystyle n\geq 2,$ we will show that it also holds for $ \displaystyle n + 1.$ Since $ \displaystyle 0\leq a_{0}\leq a_{1}\leq \cdots \leq a_{n}$ and $ \displaystyle 2a_{i}\geq a_{i - 1} + a_{i + 1}$ for all $ \displaystyle i = 1,2,\ldots ,n - 1,$ by the inductive hypothesis, we have $ \displaystyle \sum_{i = 1}^{n}a_{i}^{2}\leq \frac {4}{n(n + 1)^{2}}\left(\sum_{i = 1}^{n}ia_{i}\right) ^{2}.$ Therefore, it suffices to prove that $ \displaystyle \left[ \sum_{i = 1}^{n}ia_{i} + (n + 1)a_{n + 1}\right] ^{2}\geq \frac {(n + 1)(n + 2)^{2}}{4}\left[ \frac {4}{n(n + 1)^{2}}\left( \sum_{i = 1}^{n}ia_{i}\right)^{2} + a_{n + 1}^{2}\right] .$ Setting $ \displaystyle \sum_{i = 1}^{n}ia_{i} = \frac {n(n + 1)}{2}A,$ $ \displaystyle A\leq a_{n + 1},$ the above inequality becomes $ \displaystyle (n + 1)\left( \frac {n}{2}A + a_{n + 1}\right) ^{2}\geq \frac {(n + 2)^{2}}{4}(nA^{2} + a_{n + 1}^{2}),$ or $ \displaystyle \frac {n}{4}(a_{n + 1} - A)[(3n + 4)A - na_{n + 1}]\geq 0.$ From this, we see that the inequality for $ \displaystyle n + 1$ holds if we have $ \displaystyle A\geq \frac {n}{3n + 4}a_{n + 1},$ or $ \displaystyle \sum_{i = 1}^{n}ia_{i}\geq \frac {n^{2}(n + 1)}{2(3n + 4)}a_{n + 1}.$ Since $ \displaystyle 2a_{i}\geq a_{i - 1} + a_{i + 1}$ for all $ \displaystyle i = 1,2,\ldots ,n,$ we can easily deduce that $ \displaystyle a_{i - 1}\geq \dfrac{i - 1}{i}a_{i}$ for all $ \displaystyle i = 2,3,\ldots ,n + 1,$ and hence we get $ \displaystyle a_{i - 1}\geq \dfrac{i - 1}{i}a_{i}\geq \frac {i - 1}{i}\cdot \frac {i}{i + 1}a_{i + 1}\geq \cdots \geq \frac {i - 1}{i}\cdots \frac {n}{n + 1}a_{n + 1} = \frac {i - 1}{n + 1}a_{n + 1}.$ It follows that $ \displaystyle \sum_{i = 1}^{n}ia_{i}\geq \frac {a_{n + 1}}{n + 1}\sum_{i = 1}^{n}i(i - 1) = \frac {a_{n + 1}}{n + 1}\cdot \frac {n(n^{2} - 1)}{3} = \frac {n(n - 1)}{3}a_{n + 1}.$ Since $ \displaystyle \dfrac{n(n - 1)}{3} > \dfrac{n^{2}(n + 1)}{2(3n + 4)}$ for $ \displaystyle n\geq 2,$ we obtain $ \displaystyle \sum_{i = 1}^{n}ia_{i} > \frac {n^{2}(n + 1)}{2(3n + 4)}a_{n + 1},$ and the proof is completed.

Thank you very much for your nice proof And thanks for sharing your amazing blog

u can also use $\textbf{Chebyshev}$