An "Olympiad type" inequality:
Let $x_1,x_2,\dots,x_n$ real numbers in $[0;+\infty)$ with $x_1+\dots+x_n=1$ and
$$f(x_1,\dots,x_n)=\frac{x_1+x_2}{1+x_1x_2}+\frac{x_2+x_3}{1+x_2x_3}+\dots+\frac{x_n+x_1}{1+x_nx_1}$$ Find the minimum of $f$ under the given constraint.
It seems rather delicate to prove that $\frac95$ is the minimum for $n \geq 3$ apparently.
The problem comes from here, and as French member of the forum linked, I hope this problem will find a greater audience here, since it remains unsolved so far despite all unsuccessful attempts. Please do not submit any reasoning already rejected in the previous linked topic.
Thanks for your support.
Best Answer
Here is a complete solution proposal for the case $n\geq 4$. Sorry for the bad english.
The set of $(x_1,...,x_n)$ with non negative coefficients and sum $1$ is compact and the function $f$ considered is continuous, so we can fix $(a_1,...,a_n)$ a point where $f$ reaches its minimum.
Among all these points, we can consider one that has a maximal number of zeros in the list $(a_1,...,a_n)$.
Let us reason by contradiction and assume that this maximal number is less than or equal to $n-3$ and let $(a_1,...,a_n)$ be a associated point (which thus has at least three positive coefficients).
By circular permutation, we can assume without loss of generality that $a_1>0$ and $a_j>0$ for some $j\in [3,n-1]$.
We set $S = a_1+a_j$ and $g$ the function defined on $[0,S]$ by $g(x) = f(x,a_2,...,a_{j-1},S-x,a_{j+1},....,a_n)$.
There exists a constant $C$ independent of $x$ such that $g(x) = \dfrac{x+a_2}{1+a_2 x}+\dfrac{x+a_n}{1+a_n x} + \dfrac{a_{j-1}+S-x}{1+a_{j-1}(S-x)} + \dfrac{a_{j+1}+S-x}{1+a_{j+1}(S-x)}+C.$
Note that $j\in [3,n-1]$ allows $x$ and $S-x$ not to mix in the fractions.
For any $a\in [0,1]$, we set $h_a(x) = \dfrac{a+x}{1+ax}$ and we have $h_a''(x) = -\dfrac{2a(1-a^2)}{(1+ax)^3}\leq 0$, so each function $h_a$ is concave on $[0,S]$.
By symmetry, each function $x\mapsto h_a(S-x)$ is also concave on $[0,S]$.
By summing concave and continuous functions, the function $g$ is concave and continuous and therefore, $\min g_{[0,S]} = \min (g(0),g(S))$.
Thus, we have $\min f = f(a_1,...,a_n) = g(a_1) \geq \min (g(0),g(S)) = \min (f(0,a_2,...,a_{j-1},a_1+a_j,...,a_n), f(a_1+a_j,a_2,...,a_{j-1},0,....,a_n))$, so $f$ has a minimum at $(0,a_2,...,a_{j-1},a_1+a_j,...,a_n)$ or $(a_1+a_j,a_2,...,a_{j-1},0,....,a_n)$, which contradicts the assumed maximality of the number of zeros in $(a_1,...,a_n)$ (one of the two strictly positive terms $a_1$ or $a_j$ could have been replaced by 0 while maintaining the minimality property).
After this proof by contradiction, we know that $f$ has a minimum at some point $(a_1,...,a_n)$ with at least $n-2$ zeros.
We have $f(1,0,...,0)=2$ and $f(1/2,1/2,0,....,0) = \dfrac{9}{5}<2$, so we can affirm that the maximum number of zeros that we were interested in from the beginning is $n-2$.
Therefore, there exists a point $(a_1,...,a_n)$ such that $a_1>0$, $a_j>0$ for some $j\in [2,n]$, and $a_i=0$ for all $i\neq 1,j$ that satisfies $\min f = f(a_1,...,a_n)$.
If $j\in [3,n-1]$, by a completely similar reasoning (using the concavity argument), we have $\min f \geq \min f(a_1+a_j,a_2,..,0,a_{j+1},...,a_n) , f(0,...,a_1+a_j,...,a_n) )=f(1,0,...,0,0,...,0)=2$
which is absurd.
Therefore, by circular permutation, we can find a point $(a_1,a_2,0,...,0)$ such that $\min f = f(a_1,a_2,0,...,0)$. To conclude, we only need to minimize $x\mapsto f(x,1-x,0,...,0)$, which is achieved at $x=1/2$, showing that $\min f\geq f(1/2,1/2,0,...,0)=\dfrac{9}{5}$.