Let $x_{i}>0,i=1,2,\ldots,n$ and such $\prod_{i=1}^{n}x_{i}=x_{1}x_{2}\cdots x_{n}=1$. Show that this inequality $$(x_{1}+x_{2})(x_{2}+x_{3})\cdots (x_{n}+x_{1})\ge 2^n-n^2+n\sum_{i=1}^{n}x_{i}$$ and this inequality $$(1+\dfrac{x_{2}}{x_{1}})(1+\dfrac{x_{3}}{x_{2}})\cdots(1+\dfrac{x_{1}}{x_{n}})\ge 2^n-n^2+\dfrac{n\sum_{i=1}^{n}x_{i}}{\sqrt[n]{\prod_{i=1}^{n}x_{i}}}$$ are the same.
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Should that be $\cdots$ in your expression? – amcalde Sep 29 '14 at 14:00
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The title of this question is different from the actual question asked. IF the question given is the intended one, the answer is self-evident:
- The left-hand sides of the inequalities are equal to each other; the second one is simply obtained by dividing the first term in the product by $x_1$, second term by $x_2$ and so on, all the way until $x_n$, and we know that dividing by all these actually doesn't change anything (since the product $x_1x_2\ldots x_n$ is equal to $1$).
- The right-hand sides are almost identical, only differing in the the sum on the right-hand side being divided by $n$-th root of the same product... which is clearly equal to $1$ too.
Peter Košinár
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