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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.

River Li
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Has
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  • Seems unlikely that when $ x_1 \rightarrow 1$, we still will have $ \prod (n - x_n)/(1-x_n) \leq (n+1)^n$, so there's an error in your approach. – Calvin Lin Jan 08 '22 at 12:13
  • Maybe i should have included R \ {1} in the domain bcz x=1 breaks the right hand side of the question. – Has Jan 08 '22 at 12:22
  • That doesn't help. Take $ x_1 = 1 - (n-1)\epsilon, x_i = \epsilon$. The product goes to infinity, so it cannot be bounded above. $\quad$ I believe that you made an error in "using (3) and (2)". Can you elaborate on how you combined the inequalities? – Calvin Lin Jan 08 '22 at 12:25
  • The sum on (3) is lower bounded by the sum $\frac {n}{ \sum_{k=1}^n \frac {1}{1-x_k}}$ so decreasing the lower bound doesn't affect the inequality. By taking the max value of $\sum_{k=1}^n \frac{1}{1-x_k}$ the lower bound is just decreased. – Has Jan 08 '22 at 12:30
  • No, write that out slowly. You're skipping too many steps. $\quad$ The denominator (IE $ \sum 1/(1-x_i)$) is unbounded above, so what is the lower bound of $\prod ( 1 - x_k ) ^{1/n}$? We have "n divided by an unbounded value". In particular, convince yourself that the lower bound is 0, achieved with my example of $ x_1 = 1 - (n-1) \epsilon$. – Calvin Lin Jan 08 '22 at 12:33
  • @CalvinLin We can use tangent line method if $x_i < 1 - 1/n, \forall i$: $\ln \frac{1 - x^2}{x(n - x)} \ge \frac{-n^3}{n^2 - 1}(x - 1/n)$. Then, we deal with the remaining case. – River Li Mar 10 '22 at 15:38

1 Answers1

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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 :

$$1-c^{2}-\left(3-b\right)b\leq 0$$

$$1-a^{2}+1-c^{2}-\left(\left(3-a\right)a+\left(3-b\right)b\right)\leq 0$$

$$1-a^{2}+1-b^{2}+1-c^{2}-\left(\left(3-a\right)a+\left(3-b\right)b+\left(3-c\right)c\right)=0$$

Barackouda
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