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# Understand the problem

Let $x_1, x_2,\ldots , x_n$ be positive real numbers such that $x_1x_2\cdots x_n = 1$. Prove that $$\sum_{i = 1}^n \frac {1}{n - 1 + x_i}\le 1.$$

##### Source of the problem
Singapore Team Selection Test 2008
Inequalities
Medium
##### Suggested Book
Inequalities by BJ Venkatachala

# Start with hints

Do you really need a hint? Try it first!

Use the method of contradiction.

Suppose that $\Sigma\frac{1}{n-1+x_i}=k>1$. Then we have $latex \Sigma \frac{1}{k(n-1+x_i)}=1$.

Let’s call $\frac{1}{k(n-1+x_i)}=a_i$. We have $1=\prod x_i=\prod\left(\frac{1}{ka_i}-n+1\right)< \prod\left(\frac{1}{a_i}-n+1\right)=\prod\frac{b_i}{a_i}$ where $b_i=1-(n-1)a_i$. Hence $\prod b_i>\prod a_i$. Using the definition of $a_i$, this becomes $\prod b_i>\frac{\prod (1-b_i)}{(n-1)^n}$. To get a contradiction, find an inequality that contradicts this one.

Note that $\Sigma b_i=1$. Hence, for any given $j$, $1-b_j=\Sigma b_i-b_j\ge (n-1)\sqrt[n-1]{\frac{\prod b_i}{bj}}$. Taking product over all $j$, we get $\prod (1-b_j)\ge (n-1)^n\prod b_i$ which is a contradiction (given hint 3).

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#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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