Could you give your mathematical and logical analysis or approach about my induction proof ?
Origin of the problem This problem is Problem 2166 from Mathematics Magazine.
The only thing I don't prove here, is the result for $n=3$. The goal of this discussion is not about this case because it is much standart than the general case. I just recommand to you to read resources or to prove this case.
I hope posting here my original proof will not violate the spirit of the problem, neither the explicit rules of MAA.
I thank respectfuly MAA and the author for having submited this problem.
Problem Minimising $\frac{x_1+x_2}{1+x_1x_2}+\frac{x_2+x_3}{1+x_2x_3}+\dots+\frac{x_n+x_1}{1+x_nx_1}$ non-negative $x_i$ satisfying $x_1+x_2+\dots+x_n=1$
Claim Let $c$ a not null positive real number lower than $1$.
$\forall n \ge 3$, the minimum of $\frac{x_1+x_2}{1+x_1x_2}+\frac{x_2+x_3}{1+x_2x_3}+\dots+\frac{x_n+x_1}{1+x_nx_1}$, $x_i$ none-negative numbers satisfying $\displaystyle \sum_{i=1}^{n} x_i=c$ is $c + \frac{4 c}{4+c^2}$ for $X=X0(c)=(\frac{c}{2},\frac{c}{2},\ 0,\ \dots,\ 0)$.
Proof By induction on the number $n$ of variables $x_i$.
Base case $k=3$ is a known solved problem for which we know that the solution is for $X=(\frac{c}{2}, \frac{c}{2},\ 0) \forall c \in\ ]0;\ 1]$ that gives the minimum $c + \frac{4c}{4+c^2}$ ($\frac{9}{5}$ when $c=1$).
Induction step Let $k \gt 3$ and $(c, C) \in\ ]0;\ 1]^2$, two real numbers.
Let $X=(x_1, x_2, \dots, x_k)$, and $U=(u_1, u_2, \dots, u_{k+1})$ two finite set of real positive variables such as $x_1+x_2+\dots+x_k=c$ and $u_1+u_2+\dots+u_k+u_{k+1}=C$.
Let $g(X)=\frac{x_1+x_2}{1+x_1x_2}+\frac{x_2+x_3}{1+x_2x_3}+\dots+\frac{x_{k}+x_1}{1+x_{k}x_1}$
Let $f(U)=\frac{u_1+x_2}{1+u_1u_2}+\frac{u_2+u_3}{1+u_2u_3}+\dots+\frac{u_{k}+u_{k+1}}{1+u_{k}u_{k+1}}+\frac{u_{k+1}+u_1}{1+u_{k+1}u_1}$
We suppose that the induction hypothesis is right for $n=k$, in other words, that the minimum of $g$ is for $X=X0(c)$.
Let show in this step that it implies that it is right for $n=k+1$. In other words that the minimum of $f$ is for $U0(C)=(\frac{C}{2},\ \frac{C}{2},\ 0,\ \dots,\ 0,\ u_{k+1}=0)$.
Given that $f(U)=g(u_1,\ u_2,\ \dots,\ u_k) + \left(\frac{u_k + u_{k+1}}{1+u_k u_{k+1}} + \frac{u_{k+1} + u_1}{1+u_{k+1} u_1} - \frac{u_k + u_1}{1 + u_k u_1}\right)$ $= g(u_1,\ u_2,\ \dots,\ u_k) + h(u_k, u_{k+1}, u_1)$ with $h(u_k, u_{k+1}, u_1)=\frac{u_k + u_{k+1}}{1+u_k u_{k+1}} + \frac{u_{k+1} + u_1}{1+u_{k+1} u_1} - \frac{u_k + u_1}{1 + u_k u_1}$ $h(u_k, u_{n+1}, u_1)$ is obviously minimum for $u_{k+1}=0$, and is equal to $0$ as soon as $(u_k=0 \vee u_1=0)\ \big(h(0, 0, u_1)=u_1-\frac{u_1+0}{1+0}=u_1-u_1=0 \big)$.
So we can conclude that $f(U) \ge g(u_1,\ u_2,\ \dots,\ u_k)$.
Obviously a solution with $u_{k+1} = C$ is not possible because it would mean $u_1=u_2=\dots=u_k=0$ and $f(0,\ 0,\ \dots, C)= 2C$ which we can assert, is not the minimum.
Then $u_{k+1} \ne C$, and $(C-u_{k+1}) \gt 0$.
In other words, we have to minimise $g(U)$, under the constraint $\displaystyle \sum_{i=1}^{k} u_i=C-u_{k+1} \in\ ]0;\ 1]$.
Thus, we can use the induction hypothesis for $n=k$ and assert that $g(U)$ is minimised for $U=U0(C)=(\frac{C-u_{k+1}}{2},\ \frac{C-u_{k+1}}{2},\ 0, \dots,\ 0, u_{k+1})$
Let $m=u_{k+1}$
The calculus of $g(U0(C))$ gives :
$\dfrac{C-m}{1+\frac{(C-m)^2}{4}} +\dfrac{C-m}{2} + \frac{c+m}{2+m(C-m)}$
A study of variation of the function $d$ on $[0;\ 1]$ :
$d(x)=\dfrac{C-x}{1+\frac{(C-x)^2}{4}} +\dfrac{C-x}{2} + \frac{c+x}{2+x(C-x)}$
shows that the minimum is for $x=0$.
So $g(U)$ is minimum for $m=0=u_{k+1}$ and is equal to : $C + \frac{4 C}{4+C^2}$.
In conclusion, $U0(C)=U0=(u_1, u_2, 0,\dots, u_k=0, 0)=(\frac{C}{2}, \frac{C}{2}, 0, \dots,\ 0, u_{n+1}=0)$ minimises $f(U)$ since it minimises $g(u_1,\ u_2,\ \dots,\ u_k)$ and $h(u_k, u_{k+1}, u_1)$.
So eventually we showed that $\forall n \ge 3, \forall\ c \in\ ]0;\ 1] \frac{x_1+x_2}{1+x_1x_2}+\frac{x_2+x_3}{1+x_2x_3}+\dots+\frac{x_n+x_1}{1+x_nx_1}$ $x_i$ none-negative real numbers satisfying $x_1+x_2+\dots + x_n=c$ is minimised for $X=X0(c)=(\frac{c}{2},\frac{c}{2},\ 0,\ \dots,\ 0)$ and is equal to $g(X0(c))=c+\frac{4 c}{4 + c^2}$.
QED.
Our claim is now proved and we'll use it to solve the initial problem.
By taking $c=1$ in our claim, we proved that the minimum is $g(U0(1))=1 + \frac{4}{4+1}=\frac{9}{5}$.