$ \frac{x}{1+y^2}\ge x-\frac{xy}{2}$ Then we use the lemma $ (x_1+x_2+...+x_n)^2\ge 4(x_nx_1+x_1x_2+...+x_{n-1}x_n)$ and we're done. (Prove by induction)

When does equality hold then? By your proof, it seems no matter how many variables there are, it will always be less than a fixed value, but if we make all the variables equal, then it depends on how many variables there are.

The minimum value is 3/2, attained when $a_1 = a_2 = 1$ and $a_3 = \cdots = a_n = 0$.

dgreenb801 wrote: When does equality hold then? By your proof, it seems no matter how many variables there are, it will always be less than a fixed value, but if we make all the variables equal, then it depends on how many variables there are. I think Altheman's proof is correct. Both lemmas have equality cases $\forall x,y\in\{0,1\}$ and $a_1 = a_2 = 1,a_3 = \cdots = a_n = 0$ respectively. Nothing to do with the case of all $a_i$ equal.

Note that the lemma being used is only true for $n \ge 4$ so for n = 3 you can use smoothing to show $\frac{4}{3}$ is the minimum.

The problem requires that $n>3$ however.

Any complete and more detailed solution?

How do you find such old threads and why do you revive them?

Altheman wrote: $ \frac{x}{1+y^2}\ge x-\frac{xy}{2}$ Then we use the lemma $ (x_1+x_2+...+x_n)^2\ge 4(x_nx_1+x_1x_2+...+x_{n-1}x_n)$ and we're done. (Prove by induction) Could also straight up use a two-step smoothing process to find maximum of $a_1a_2 + a_2a_3 + \dots a_na_1$: Suppose $n \geq 5$. Else it is trivial. i) Push an element to 0: WLOG assume $a_1 + a_3 \geq a_3 + a_5$. Then push $a_4$ to $a_2$. ii) Apply chain reaction to push all irrelevant elements to 0: Reorder elements to $a_1a_2 + a_2a_3 + \dots a_{k-1}a_{k}$ Push $a_1$ to $a_3$ and repeat (reordering to make indices nicer) until we have $a_1a_2 + a_2a_3$ left, at which point it becomes trivial.

Let $a_1,a_2,\cdots,a_n$ $(2\leq n\leq 4)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove that $$\frac{a_1}{a^2_2+1}+\frac{a_2}{a^2_3+1}+\cdots+\frac{a_{n-1}}{a^2_n+1}+ \frac{a_n}{a^2_1+1}\geq\frac{n}{2}.$$Let $a_1,a_2,\cdots,a_n$ $( n\geq 5)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove that $$\frac{a_1}{a^2_2+1}+\frac{a_2}{a^2_3+1}+\cdots+\frac{a_{n-1}}{a^2_n+1}+ \frac{a_n}{a^2_1+1}>\max\{\frac{8n}{n^2+8} ,n-\frac{n^2}{8}\}.$$(Zhangyanzong)

Let $a_1,a_2,\cdots,a_n$ $( n\geq 2)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Find the minimum possible value of $$\frac{a_1}{a^3_2+4}+\frac{a_2}{a^3_3+4}+\cdots+\frac{a_{n-1}}{a^3_n+4}+

peregrinefalcon88 wrote: Note that the lemma being used is only true for $n \ge 4$ so for n = 3 you can use smoothing to show $\frac{4}{3}$ is the minimum. https://artofproblemsolving.com/community/c6h1820982p12167038: Let $a, b, c$ be nonegative real numbers such that $a + b + c = 2$. Then$$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq\frac{4}{3}$$Proof of DievilOnlyM:$$2-\sum\frac{a}{1+b^2}=\sum\frac{ab^2}{1+b^2}\leq\sum\frac{ab^2}{2b}=\frac{1}{2}\sum ab\leq\frac{2}{3}$$

https://dgrozev.wordpress.com/2020/04/16/%D0%B0-weaker-jensens-inequality-an-olympiad-approach/

dgrozev wrote: https://dgrozev.wordpress.com/2020/04/16/%D0%B0-weaker-jensens-inequality-an-olympiad-approach/ splendid extension!

The maximum is $\frac 32$, achieved at $(1, 1, 0, 0, \dots, 0)$. The key is to use the tangent line at $x=1$, which gives us the estimate $$\frac 1{x^2+1} \geq 1-\frac x2 \iff \frac{x(x-1)^2}{2(x^2+1)} \geq 0.$$Thus, $$\sum_{i=1}^n \frac{a_i}{a_{i+1}^2+1} \geq \sum_{i=1}^n a_i - \frac 12 a_ia_{i+1} \geq \frac 32.$$

We want to show that the minimum is $\frac{3}{2}$, which is attainable when all of the terms are $0$ except for two consecutive ones with value $1$. First, define $f:=\dfrac{1}{x^2+1}$.We now have (tangent line trick) : $$f(x) \geq \frac{-x}{2}+1 \iff \frac{(x-1)^2x}{2(x^2+1)} \geq 0 \quad \forall x \in \mathbb{R}$$ Hence: $$\sum \frac{a_i}{1+a_{i+1}} \geq \sum a_i - \frac{a_ia_{i+1}}{2} \geq \frac{3}{2}$$ Where the last inequality comes $$\sum a_ia_{i+1} \leq \sum a_ia_j \leq \frac{1}{4} \left( \sum a_i \right)^2 $$ $$\mathbb{Q.E.D.}$$

We claim that the minimum possible value is $\boxed{\frac{3}{2}}$. We can see that this is given by $(1, 1, 0, 0, \ldots , 0)$. We claim that \[\frac{1}{a^2+1} \geq \frac{-a+2}{2}\]for $a \in [0, 2]$. Multiplying both sides by $2a^2+2$ we get \[a^3-2a^2+a \geq 0 \Rightarrow a(a+1)^2 \geq 0\]which is obvious for $a \in [0, 2]$. Now we can see that since $\frac{a_n}{a_{n+1}^2+1} \geq a_n - \frac{a_na_{n+1}}{2}$ We simply need \[\sum_{\text{cyc}} a_1a_2 \leq 1\]We claim that \[4 \sum_{\text{cyc}} a_1a_2 \leq (a_1+a_2+\dots+a_n)^2\]Which obviously proves the problem. This last inequality is true because consider $(a_i, a_{i+2})$ we can see that we can increase the sum the most if we let one of these pair be $0$. So set this to either $(a_i+a_{i+2}, 0)$ or $(0, a_i+a_{i+2})$ (based on which of $a_{i-1}$ and $a_{i+3}$ is bigger). Thus, once we do this enough times for $0 \pmod 2$ and $1 \pmod2$ indices we will get only zero indices besides two indices. Thus we need that $4xy \leq (x+y)^2$ which is obvious. $\blacksquare$ For some reason many people are omitting the very untrivial proof of the last lemma