instinct tells me cauchy will work (i'm trying the prb neway)

apply cauchy schwarz to \sum [(a_i)^2*{(a_(i+1)^2+1] and \sum [ (a_i)/{(a_(i+1)^2 + 1] and multiple these two to get their product greater than the rhs of the inequality.so it suffices to prove \sum [(a_i)^2*{(a_(i+1)^2+1] <= 5/4 which is equivalent to \sum [ (a_i)^2*{a_(i+1)}^2] <= 1/4 denote by x_i = (a_i)^2 and one has to prove x_1*x_2 +x_2*x_3 + ..... + x_n*x_1 <= 1/4. now i havent figured this out clearly but wot i did was to fix n-2 of the x_i's and vary the other 2 and get some conditions for maxima.check that method and give me a sign if u need more help or if something is wrong with my method

I know two methods of solving this inequality (one of them I found during the actual exam) and another was told to me by someone else. When you want a solution, let me know.

I guess that the more solutions we have the better it is, because perhaps one of them will work on another problem.. so Valentin Vornicu.. I hope you won't mind posting your solutions! thank you!

Here's my solution(although I participated in that exam and I got 0 for this question ): \sum (ai/(a(i+1)^2+1))= \sum (ai^3/(a(i+1)^2*ai^2+ai^2). By using the inequality (easy to prove, just prove it for 2 numbers and then extend) \sum (ai^2/xi)>= (\sum ai)^2/ \sum xi where ai and xi are positive numbers, we can prove that the left part is>= ( \sum ai^3/2)^2/ \sum (a(i+1)^2*ai^2+ai^2). To prove that this one, in turn, is >= 4/5 etc, etc, we get to the same thing as you found, that is: if \sum xi=1 then \sum xi*x(i+1)<= 1/4. Here's how we can prove this: For n even we have \sum xi*x(i+1)<= ( \sum x(2i))* (\sum x(2i-1))<= 1/4 from AM-GM. Just do something similar for n odd. Hope it's correct!

My solution also involves Cauchy: We have 4/5*(a1 \sqrt a1+...+an \sqrt an)^2 \leq 4/5*(a1^2+...+an^2)(a1+...+an)=4/5*(a1+....+an) We also have a1/(a2^2+1)+....+an/(a1^2+1)>=n(a1+...+an)/(a1^2+..+an^2+n)=n(a1+...+an)/(1+n) So we now have : n(a1+....+an)/(1+n)>=4/5(a1+....+an) n/(1+n)>=4/5 5n>=4(1+n) n>=4 which is true

I think my sol for the inequality is wrong. Sorry!

n>4: let f(x_1,x_2,..,x_n)=x_1*x_2+x_2*x_3+..+x_n*x_1 You cann also notice that if x_n=min(x_1,x_2,..,x_n) then f(x_1,x_2,..,x_(n-1)+x_n)>=f(x_1,x_2,..,x_n). f(x_1,x_2,..,x_(n-1)+x_n)-f(x_1,x_2,..,x_n) =[x_(n-2)+x_1]*[x_(n-1)+x_n] - x_(n-2)*x_(n-1)-x_(n-1)*x_n-x_n*x_1 =x_(n-2)*x_n + x_(n-1)*(x_1-x_n) >= 0 because x_1>=x_n Hence, to prove f(x_1,x_2,..,x_n)<=4 we just have to prove for n=4, which is not difficult.

The $LHS$ can be rewritten as $\frac{a_1^3}{a_1^2a_2^2+1} + \cdots + \frac{a_n^3}{a_n^2a_1^2+1}$ By Titus Lemma it is $\geqslant \frac{\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2}{1+a_1^2a_2^2+\cdots +a_n^2a_1^2} $ Simplifying it remains to prove that $\frac{1}{4} \geqslant a_1^2a_2^2+ \cdots +a_n^2a_1^2$ which is true by chebyshev as $n\geqslant4$

Let $n=2,3$ , and let $a_1,a_2,\ldots,a_n$ be positive real numbers such that \[ a_1^2+a_2^2+\cdots +a_n^2=1 . \]Prove that the following inequality takes place \[ \frac{a_1}{a_2^2+1}+\cdots +\frac{a_n}{a_1^2+1} \geq \frac{n}{n+1}\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2 . \]