# Prove that recursive sequence is greater than 3/11

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 $n\geq 2$. – Jonas Meyer Nov 15 '12 at 3:45

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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I'm guessing for the minimum, you take the first derivative and solve for it when it equals zero. I don't see how I can solve it. Thanks. – Sunny Brooke Nov 15 '12 at 4:11
nvm I see it. Wow this solution is much simpler than I would have expected. Are you sure we can do this by considering the sequence as a function and finding the minimum? Thanks. – Sunny Brooke Nov 15 '12 at 4:16
yes. Every term $x_n$ in the sequence is the evaluation of the function at some positive point, $x_{n-1}$, to be precise. So, none of them can be smaller than the minimum of the function. – dexter04 Nov 15 '12 at 4:41
Ingenious! Thank you so much. – Sunny Brooke Nov 15 '12 at 5:13