Let $x_1,x_2,……,x_n$ be positive real numbers such that $x_1+x_2+x_3+….+x_n=1$. Prove that $$\prod_{k=1}^n \frac{1+x_k}{x_k} \geq \prod_{k=1}^n \frac{n-x_k}{1-x_k}.$$
My approach using Am-Gm inequality $$\sum_{k=1}^n \frac{n-x_k}{n} \geq \prod_{k=1}^n(n-x_k)^\frac {1}{n}$$ which implies $$\prod_{k=1}^n (n-x_k) \leq (\frac {n^2-1}{n})^n. \tag{1}$$
By using Am-Hm inequality$$ \sum_{k=1}^n(1-x_k).\sum_{k=1}^n \frac{1}{1-x_k} \geq n^2$$ which implies $$\sum_{k=1}^n \frac{1}{1-x_k} \geq \frac{n^2}{n-1}. \tag{2}$$
Further using Gm-Hm $$\prod_{k=1}^n (1-x_k)^ \frac{1}{n}\geq \frac {n}{ \sum_{k=1}^n \frac {1}{1-x_k}}. \tag{3} $$
Using (3) and (2), we get$$\prod_{k=1}^n (1-x_k) \geq (\frac{n-1}{n})^n$$ Combining the above result with (1) we get $$\prod_{k=1}^n \frac{n-x_n}{1-x_n} \leq (n+1)^n$$ Similarly I proved $$ \prod_{k=1}^n \frac{1+x_k}{x_k}\geq (n+1)^n$$My question is is there a better way to approach this and is my approach even correct? Any help shall be appreciated.
Best Answer
Hint :
Use Karamata's inequality and the convexity of $-\ln(x)$ on $(0,\infty)$.
By example in the case $n=3$ for $b\geq a \geq c>0$ and $a+b+c=1$ we have :
$$1-c^{2}-\left(3-b\right)b\leq 0$$
$$1-a^{2}+1-c^{2}-\left(\left(3-a\right)a+\left(3-b\right)b\right)\leq 0$$
$$1-a^{2}+1-b^{2}+1-c^{2}-\left(\left(3-a\right)a+\left(3-b\right)b+\left(3-c\right)c\right)=0$$