VMO inequality

Algebra Level 4

$\displaystyle\sum_{i=1}^{27} \frac{1}{x_i+1}=1$

Let $x_1, x_2,\ldots ,x_{27}$ be positive reals that satisfy the above condition. If the minimum value of $\displaystyle\prod_{i=1}^{27} x_i$ can be written as $a^b$ , where $a$ and $b$ are positive integers with $a$ cube-free, find $a+b$ .

Bonus Generalize for $x_i \ (i=\overline{1,n})$ .

The answer is 53.

1 solution

Guillermo Templado
Dec 17, 2016

Let's call $x_i + 1 = y_i$ , then applying AM - GM inequality we have: $\displaystyle \sqrt[27]{\prod_{i = 1}^{27} \frac{1}{y_i}} \leq \sum_{i = 1}^{27} \frac{\frac{1}{y_i}}{27} = \frac{1}{27}$ powering to 27 we get the inequality $\prod_{i = 1}^{27} \frac{1}{y_i} \leq (\frac{1}{27})^{27} \Rightarrow 27^{27} \leq \prod_{i = 1}^{27} y_i$ Now, making $y_1 = y_2 = ... = y_{27} = 27 \Rightarrow x_1 = x_2 = ... = x_{27} = 26$ we get the equality $27^{27} = \prod_{i = 1}^{27} y_i \Rightarrow 26^{27} = \prod_{i = 1}^{27} x_i$ where we get the minimum value for this product, so $a + b = 26 + 27 = 53$ .

For making the bonus we can use the same process...

VMO inequality

The answer is 53.

1 solution

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