All sums in the solution below are cyclic. Lemma: $\displaystyle \sum \dfrac{x_1}{x_1+x_2} \geq \dfrac{n}{2}$. Proof: We prove the desired inequality with induction on $n$. For $n=3$ it is equivalent to $(x_1-x_2)(x_2-x_3)(x_1-x_3) \geq 0,$ which is true since $x_1 \geq x_2 \geq x_3$. Assume the inequality is true for $n$ variables. Consider the sum as a function $f$ of $x_1$. Its derevative is $f'(x_1)=\dfrac{x_2}{(x_1+x_2)^2}-\dfrac{x_n}{(x_n+x_1)^2}$. Note that $f'(x_1)=\dfrac{(x_2-x_n)(x_1^2-x_2x_n)}{(x_1+x_2)^2(x_n+_1)^2} \geq 0,$ since $x_1 \geq x_2 \geq x_n$. Therefore, $f$ is increasing, and so $\displaystyle f(x_1) \geq f(x_2)=\dfrac{1}{2}+\sum \dfrac{x_2}{x_2+x_3},$ and so we are reduced to the case of $n-1$ variables ($x_2 \geq x_3 \geq \ldots \geq x_{n-1}$), and so we are done by the inductive hypothesis $\blacksquare$ To the problem, consider the LHS divided by $k+1$ as a function $g$ in $k$. Its derivative is $g'(k)=\dfrac{1}{(k+1)^2} \sum \dfrac{x_2-x_1}{x_2+x_3}$. If $g'$ is nonnegative, then it suffices to prove the inequality only in the case of $k=1$, which is an easy application of AM-GM for $n$ terms. If $g'$ is negative, though, then we obtain that $\displaystyle \sum \dfrac{x_1}{x_2+x_3} \geq \sum \dfrac{x_2}{x_2+x_3},$ and so to prove the desired inequality we are left to prove that $\displaystyle \sum \dfrac{(k+1)x_2}{x_2+x_3} \geq \dfrac{n(k+1)}{2},$ or equivalently that $\displaystyle \sum \dfrac{x_1}{x_1+x_2} \geq \dfrac{n}{2},$ which follows from the Claim, and so we are done.

Funny problem. Note that $\sum \frac{x_1+kx_2}{x_2+x_3}=\sum \frac{x_1+x_2}{x_2+x_3}+\sum \frac{(k-1)x_2}{x_2+x_3}\ge n+(k-1)(\sum \frac{x_2}{x_2+x_3})$ by AMGM. It suffices to show $\sum \frac{x_1}{x_1+x_2}\ge \frac{n}{2}$. We use induction with $n=3$ equivalent to $(x_1-x_2)(x_1-x_3)(x_2-x_3)\ge 0$. Suppose true for $n-1$. It suffices to show $\frac{x_1}{x_1+x_2}+\frac{x_2}{x_2+x_3}\ge \frac{1}{2}+\frac{x_1}{x_1+x_3}$ which is equivalent to $(x_2-x_3)(x_1-x_3)(x_1-x_2)\ge 0$ as well. Hence we are done.

This problem was proposed by Ilija Jovcheski.

oVlad wrote: Let $k > 1{}$ be a real number, $n\geqslant 3$ be an integer, and $x_1 \geqslant x_2\geqslant\cdots\geqslant x_n$ be positive real numbers. Prove that \[\frac{x_1+kx_2}{x_2+x_3}+\frac{x_2+kx_3}{x_3+x_4}+\cdots+\frac{x_n+kx_1}{x_1+x_2}\geqslant\frac{n(k+1)}{2}.\]Ilija Jovcheski Doesnt $k$ just need to be $\geq 0$ ?

Notice that $$\frac{x_1+kx_2}{x_2+x_3}+\frac{x_2+kx_3}{x_3+x_4}+\cdots+\frac{x_n+kx_1}{x_1+x_2}$$Is equal to the sum of these two expressions $$\frac{x_1+x_2}{x_2+x_3} + \frac{x_2+x_3}{x_3+x_4}+ \cdots + \frac{x_n+x_1}{x_1+x_2}$$$$ (k-1) \left(\frac{x_2}{x_2+x_3}+\frac{x_3}{x_3+x_4}+\cdots+\frac{x_1}{x_n+x_1}\right)$$By $AM-GM$, $$(*) \ \ \ \frac{x_1+x_2}{x_2+x_3} + \frac{x_2+x_3}{x_3+x_4}+ \cdots + \frac{x_n+x_1}{x_1+x_2} \ge n\sqrt[n]{\frac{(x_1+x_2)(x_2+x_3)\cdots(x_n+x_1)}{(x_2+x_3)(x_3+x_4)\cdots(x_1+x_2)}}=n$$Since $k>1$ and $x_1 \ge x_2\ge\cdots\ge x_n$, therefore $$(**) \ \ \ (k-1)\left(\frac{x_2}{x_2+x_3}+\frac{x_3}{x_3+x_4}+\cdots+\frac{x_1}{x_n+x_1}\right)\ge (k-1)\left(\frac{x_2}{x_2+x_2}+\frac{x_3}{x_3+x_3}+\cdots+\frac{x_1}{x_1+x_1}\right)=\frac{n(k-1)}{2}$$Summing $(*)$ and $(**)$ yields the desired conclusion.

$$ \frac{x_1 + kx_2}{x_2 + x_3} + ... + \frac{x_n + kx_1}{x_1+x_2} = \frac{x_1 + x_2}{x_2 + x_3} + ... + \frac{x_n + x_1}{x_1 + x_2} + (k-1) ( \frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1+x_2}) \ge $$$$ \ge n + (k-1)(\frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1 + x_2}) \ge n + (k-1)(\frac{x_2}{2 \cdot x_2} + ... + \frac{x_1}{2 \cdot x_1} = n + (k-1) \frac{n}{2} = \frac{n(k+1)}{2} $$

Hertz wrote: $$ \frac{x_1 + kx_2}{x_2 + x_3} + ... + \frac{x_n + kx_1}{x_1+x_2} = \frac{x_1 + x_2}{x_2 + x_3} + ... + \frac{x_n + x_1}{x_1 + x_2} + (k-1) ( \frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1+x_2}) \ge $$$$ \ge n + (k-1)(\frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1 + x_2}) \ge n + (k-1)(\frac{x_2}{2 \cdot x_2} + ... + \frac{x_1}{2 \cdot x_1} = n + (k-1) \frac{n}{2} = \frac{n(k+1)}{2} $$ Xn/2Xn ???? Is not true.

b_behruz wrote: Hertz wrote: $$ \frac{x_1 + kx_2}{x_2 + x_3} + ... + \frac{x_n + kx_1}{x_1+x_2} = \frac{x_1 + x_2}{x_2 + x_3} + ... + \frac{x_n + x_1}{x_1 + x_2} + (k-1) ( \frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1+x_2}) \ge $$$$ \ge n + (k-1)(\frac{x_2}{x_2 + x_3} + ... + \frac{x_1}{x_1 + x_2}) \ge n + (k-1)(\frac{x_2}{2 \cdot x_2} + ... + \frac{x_1}{2 \cdot x_1} = n + (k-1) \frac{n}{2} = \frac{n(k+1)}{2} $$ Xn/2Xn ???? Is not true. We are maximizing the denominator so we are getting a smaller number. I don't see an issue

Note that \[\sum_{i = 1}^{n}\frac{x_i + kx_{i + 1}}{x_{i + 1} + x_{i + 2}} = \sum_{i = 1}^{n}\frac{x_{i} + x_{i + 1}}{x_{i + 1} + x_{i + 2}} + (k - 1)\sum_{i = 1}^{n}\frac{x_{i}}{x_{i} + x_{i + 1}},\]where indices are taken modulo $n$. By AM-GM we have \[\sum_{i = 1}^{n}\frac{x_{i} + x_{i + 1}}{x_{i + 1} + x_{i + 2}} \geq n,\]so it suffices \[(k - 1)\sum_{i = 1}^{n}\frac{x_{i}}{x_{i} + x_{i + 1}} \geq (k - 1)\frac{n}{2} \Longleftrightarrow \sum_{i = 1}^{n}\frac{x_{i}}{x_{i} + x_{i + 1}} \geq \frac{n}{2}.\]Now, define $a_i = x_{i + 1}/x_i$, so $a_i \leq 1$ for $1 \leq i \leq n - 1$ and $a_1a_2\cdots a_n = 1$. We have to prove $\sum_{1 \leq i \leq n} \frac{1}{1 + a_i} \geq n/2$. Claim. We have

If $x_1, \ldots, x_{n - 1} \leq 0$ are such that $a_i = e^{x_i}$, we know that $\{x_1, x_2, \ldots, x_{n - 1}\} \prec \{x_1 + \cdots + x_{n - 1}, 0, \ldots, 0\}$, so Karamata's inequality says \[\sum_{i = 1}^{n - 1}\frac{1}{1 + e^{x_i}} = \sum_{i = 1}^{n - 1}f(x_i) \geq (n - 2)f(0) + f(x_1 + \cdots + x_{n - 1}) = \frac{n - 2}{2} + \frac{1}{1 + e^{x_1 + \cdots + x_{n - 1}}},\]which proves the claim. To finish, we have \[\sum_{i = 1}^{n}\frac{1}{1 + a_i} \geq \frac{n - 2}{2} + \frac{1}{1 + a_1\cdots a_{n - 1}} + \frac{1}{1 + a_n} = \frac{n - 2}{2} + \frac{1}{1 + 1/a_n} + \frac{1}{1 + a_n} = \frac{n}{2}.\]