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We just started learning sequences and my teacher gave us this problem that seems to be incredibly hard. I don't even know where to start.

Question: The first term of the following sequence is $1$.

$ \left\{x_{n+1}\right\}^\infty_{n=1}=\left\{\frac{x^5_n + 1}{5x_n}\right\}^\infty_{n=1} $

Show that $x_n>\frac{3}{11}$, for $n\geq1$.

Any idea? Thanks.

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    You can show by induction that $\dfrac{2}{5}\leq x_n\leq \dfrac{3}{5}$ for all $n\geq 2$.2012-11-15

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First note that $x_n > 0 $ for all $n$. This fact is simple to prove.

So, all $x_n$ is of the form $x_n = \frac{x_{n-1}^5 +1}{5x_{n-1}}$

Now, consider the function $f(x) = \frac{x^5+1}{5x}$ By basic calculus, you can show that the minimum value of this function for positive $x$ is $2^{-1.6} \approx 0.33$. Also, $\frac{3}{11}=0.272727\ldots$.

So, for all $n\geq 1$, $x_n \geq 0.33 >\frac{3}{11}$

Hence proved.

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    Ingenious! Thank you so much.2012-11-15