orl wrote: Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 + a^2_2 + \ldots + a^2_{100} = 1.$ Prove that \[ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}. \] This is not a neat problem. In fact, the official solution proves the stronger inequality with the constant $ \frac{\sqrt{2}}{3}$.

hmmm, With Cauchy and Chebychev I can get it down to $ \frac{\sqrt{3}}{3}$

orl wrote: Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 + a^2_2 + \ldots + a^2_{100} = 1.$ Prove that \[ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}. \] Author: Marcin Kuzma, Poland Anybody has idea ?

use Lagrangre Multiplier method,I found The Maximum of S (our RHS) is $ 0,458879$

Allnames wrote: use Lagrangre Multiplier method,I found The Maximum of S (our RHS) is $ 0,458879$ would you post the full solutions?

Any body have datails?

Can someone please post a solution properly for this question? I am in desperate neeed of it.

Allnames wrote: use Lagrangre Multiplier method,I found The Maximum of S (our RHS) is $ 0,458879$ Are you sure?

who can show me When does Equalities hold ?

tin_124 wrote: who can show me When does Equalities hold ? It doesn't hold.

Could anyone post the solution or even a sketch of it because it's rather hard inequality and if I'm right this problem has been rejected by IMO Commete. I can't prove it myself but I'm very interested in the method which kills it.

Without knowing Lagrange multipliers this problems is nearly impossible. Let $\sum a_i^2a_{i+1} =S$. Using Lagrange multipliers, one finds that the maximum of $S$ is attained if $a_{i-1}^2+2a_ia_{i+1}=-2\lambda a_i$ for all $i$. Since $\lambda$ is constant, the sequences $(a_{i-1}^2+2a_ia_{i+1})$ and $(a_i)$ are the same up to some constant factor. Hence one can use Cauchy-Schwarz as follows: \[(3S)^2=\left(\sum a_i(a_{i-1}^2+2a_ia_{i+1}) \right)^2 \le \left( \sum a_i^2 \right) \left( \sum (a_{i-1}^2+2a_ia_{i+1})^2 \right) \mbox{.}\] From here it's not hard to obtain the bound $\sqrt{2}/3$.

Solution : Let $ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 = S $ Applying the Cauchy-Schwarz inequality to sequences $(a_{k+1})$ and $(a_{k}^2 + 2a_{k+1}+2a_{k+2})$ , and then the AM-GM inequality to numbers $a_{k+1}^2$ and $a_{k+2}^2$. (As usual, we consider the indices modulo 100, e.g. We set $a_{101} = a_{1}$ and $a_{102} = a_{2}$.) \[ (3S)^2 = ( \sum_{k=1}^{100} a_{k+1}(a_{k}^2 + 2a_{k+1}a_{k+2}) )^2 \le ( \sum_{k=1}^{100} a_{k+1}^2)( \sum_{k=1}^{100} (a_{k}^2 + 2a_{k+1}a_{k+2})^2) \]\[ = 1. ( \sum_{k=1}^{100} (a_{k}^2 + 2a_{k+1}a_{k+2}))^2 = \sum_{k=1}^{100} (a_{k}^4 + 4a_{k}^2a_{k+1}a_{k+2}+4a_{k+1}^2a_{k+2}^2) \]\[ \le \sum_{k=1}^{100} (a_{k}^4 + 2a_{k}^2(a_{k+1}^2+a_{k+2}^2) + 4a_{k+1}^2a_{k+2}^2) = \sum_{k=1}^{100} (a_{k}^4 + 6a_{k}^2a_{k+1}^2 + 2a_{k}^2a_{k+2}^2) \] Now , applying the trivial estimates : \[ \sum_{k=1}^{100} (a_{k}^4 + 2a_{k}^2a_{k+1}^2 + 2a_{k}^2a_{k+2}^2) \le ( \sum_{k=1}^{100} a_{k}^2 ) \]\[ ( \sum_{k=1}^{100} a_{k}^2a_{k+1}^2 ) \le ( \sum_{i=1}^{50} a_{2i-1}^2 )( \sum_{j=1}^{50} a_{2j}^2) \] We obtain : \[ (3S)^2 \le ( \sum_{k=1}^{100} a_{k}^2)^2 + 4( \sum_{i=1}^{50} a_{2i-1}^2 )( \sum_{j=1}^{50} a_{2j}^2) \]\[ \le 1 + (\sum_{i=1}^{50} a_{2i-1}^2 + \sum_{j=1}^{50} a_{2j}^2)^2 = 2 \] So : \[ S = a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 \le \frac {\sqrt {2}}{3} < \frac {12}{25} \]

Nice solution

Let $x_1,x_2,\cdots,x_n (n\ge 3) $ be non-negative real numbers satisfying $x_1+x_2+\cdots+x_n=1$ ,Show that\[x^2_1x_2+x^2_2x_3+\cdots+x^2_{n-1}x_n+x^2_nx_1\leq \frac {4}{27}.\]

msceok wrote: orl wrote: Let $ a_1, a_2, \ldots, a_{100}$ be nonnegative real numbers such that $ a^2_1 + a^2_2 + \ldots + a^2_{100} = 1.$ Prove that \[ a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}. \] This is not a neat problem. In fact, the official solution proves the stronger inequality with the constant $ \frac{\sqrt{2}}{3}$. please enter ur solution. its right. because 12/25>ur solution.

orl wrote: Author: Marcin Kuzma, Poland PhD Marcin KUCZMA

Simply just cauchy!!

I hope this is correct For the sake of a clearer and more legible solution, let $a_1^2=x_1\dots a_{100}^2=x_{100}, \text{ and, }a_1=y_1\dots a_{100}=y_{100}$, thus the condition becomes $\sum_{i=1}^{100}x_i=1 \text{, and } \sum_{i=1}^{100}x_iy_{i+1}<\frac{12}{25}$ Furthermore notice that: $\sum_{i=1}^{100}x_iy_{i+1}\overset{\text{Chebyshev}}{\le}100\bigg(\frac{1}{100}\sum_{i=1}^{100}x_i\bigg)\bigg(\frac{1}{100}\sum_{i=1}^{100}y_i\bigg)=\sum_{i=1}^{100}\frac{1}{100}y_i$ Now, $\sum_{i=1}^{100}\frac{1}{100}y_i\overset{\text{C-S}}{\le}\sqrt{\sum_{i=1}^{100}y_i^2}\sqrt{\frac{1}{100}}$, however notice that $\sum_{i=1}^{100}y_i^2=\sum_{i=1}^{100}x_i=1$, thus the inequality is equivalent to $\frac{1}{10}<\frac{12}{25}$, which is clearly true $\blacksquare$

We apply the method of Lagrange Multipliers. Let $f(a_1,\dots,a_{100})=a^2_1 a_2 + a^2_2 a_3 + \dots + a^2_{100} a_1$ and $g(a_1,\dots,a_{100})=a^2_1 + a^2_2 + \dots + a^2_{100}$. We have $\nabla f = \lambda \nabla g$ and $g=0$. Since $f_{a_i}=2a_ia_{i+1}+a_{i-1}^2$ and $g_{a_i}=2a_i$, we have that at the maximum of $f$, we must have \[2a_ia_{i+1}+a_{i-1}^2=\lambda 2a_i\]for all $i$. Thus, $2a_i^2a_{i+1}+a_{i-1}^2a_i=\lambda 2a_i^2$ and adding this up for $i=1,\dots,100$ gives $3f=2\lambda g=2\lambda$. By Cauchy Schwarz, \[4\lambda^2 = \left(\sum_{i=1}^{100}{2\lambda a_i^2}\right)^2=\left(\sum_{i=1}^{100}(a_i)(2a_ia_{i+1}+a_{i-1}^2)\right)^2\le \sum_{i=1}^{100}(2a_ia_{i+1}+a_{i-1}^2)^2\]Now, we'll expand this sum and apply some inequalities. We have \begin{align*} \sum_{i=1}^{100}(2a_ia_{i+1}+a_{i-1}^2)^2 &= \sum_{i=1}^{100}(4a_i^2a_{i+1}^2+a_{i-1}^4+4a_{i-1}^2a_ia_{i+1}) \\ &\le \sum_{i=1}^{100}(4a_i^2a_{i+1}^2+a_{i-1}^4+2a_{i-1}^2(a_i^2+a_{i+1}^2)) \\ &= \sum_{i=1}^{100}(2a_i^2a_{i+1}^2+a_{i-1}^4+2a_{i-1}^2a_{i+1}^2)+\sum_{i=1}^{100}(4a_i^2a_{i+1}^2) &\le 2\left(\sum_{i=1}^{100}a_i^2\right)^2 = 2 \end{align*}Thus, $2\lambda\le \sqrt{2}$ and so $f\le \frac{\sqrt2}{3}<0.48$

F10tothepowerof34 wrote: I hope this is correct You can't apply Chebyshev since the sequences aren't sorted

@below Thanks for the correction

PEKKA wrote: Note: the bound I get is very weird and unlike everyone else so if I got something wrong pls lmk. This implies $$(a_1^4+a_2^4+\dots a_{100}^4)(a_2^2+a_3^2+\dots a_n^2+a_1^2)\ge (\frac{1}{100})(1) \ge (\sum_{cyc}a_1^2a_2)^2. $$ Why $\frac{1}{100} \ge (\sum_{cyc}a_1^2a_2)^2$? We know only that $(a_1^4+a_2^4+\dots a_{100}^4)(a_2^2+a_3^2+\dots a_n^2+a_1^2) \geq \frac{1}{100}$ and $(a_1^4+a_2^4+\dots a_{100}^4)(a_2^2+a_3^2+\dots a_n^2+a_1^2) \geq (\sum_{cyc}a_1^2a_2)^2$, not something more.