Let $n \geq 3$ be an integer and $k>1$ be a real number.

Question

Let $n \geq 3$ be an integer and $k>1$ be a real number. Consider the sequence $x_1 \geq x_2 \geq x_3 \geq .... \geq x_n$ be positive real numbers. Prove that $ \dfrac{x_1+kx_2}{x_2+x_3}+\dfrac{x_2+kx_3}{x_3+x_4}+....+\dfrac{x_n+kx_1}{x_1+x_2} \geq \dfrac{n(k+1)}{2}$

Sadly I haven't been able to find any approach regarding the question except the fact that I tried isolating each term on left hand side and compared it individually to $\frac{(k+1)}{2}$ hoping to add up all the inequalities but just met with the problem that for the last term and before that the inequality is no more valid.
I tried a method of contradiction but my proof seemed invalid as in assuming each statement is false and adding them up to get something which should be false but is actually true.
But again the sequential inequality wasn't used so I'm doubtful of it.

That's all I've tried yet and I am a beginner in inequality so I kindly request someone to help me with it.

Answer 1

By the Calvin Lin's hint: $$\sum_{cyc}\frac{x_1+kx_2}{x_2+x_3}=\sum_{cyc}\frac{x_1+x_2}{x_2+x_3}+(k-1)\sum_{cyc}\frac{x_2}{x_2+x_3}\geq n+(k-1)\sum_{cyc}\frac{x_2}{x_2+x_3}$$ and it's enough to prove that: $$\sum_{cyc}\frac{x_1}{x_1+x_2}\geq\frac{n}{2},$$ which is true by induction.

For example for $n=4$ we need to prove that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq2,$$ where $a\geq b\geq c\geq d>0$.

Indeed, $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{3}{2}$$ is true because it's just the Schur's non-negative polynomial: $$\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c)\geq0 .$$ Thus, $$\sum_{cyc}\frac{a}{a+b}\geq\frac{3}{2}-\frac{c}{c+a}+\frac{c}{c+d}+\frac{d}{d+a}=\frac{1}{2}+\frac{a}{a+c}+\frac{c}{c+d}+\frac{d}{d+a}\geq\frac{1}{2}+\frac{3}{2}=2.$$ Can you end it now?