We are tasked with proving that $a_n^2 > 3$ for all $n$ in the given sequence defined by:

$a_0 = 2, \quad a_{n+1} = \frac{2a_n}{3} + \frac{1}{a_n}.$

Step 1: Verify the inequality for the base case $n = 0$

For $n = 0$, we know $a_0 = 2$. Computing $a_0^2$, $a_0^2 = 2^2 = 4 > 3.$ Thus, the inequality holds for $n = 0$.

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Step 2: Induction hypothesis

Assume that the inequality holds for some $n = k$, i.e., $a_k^2 > 3.$

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Step 3: Prove $a_{k+1}^2 > 3$

From the recurrence relation, $a_{k+1} = \frac{2a_k}{3} + \frac{1}{a_k}$. We need to analyze $a_{k+1}^2$.

1. Compute $a_{k+1}^2$: $a_{k+1}^2 = \left( \frac{2a_k}{3} + \frac{1}{a_k} \right)^2.$

Expand the square: $a_{k+1}^2 = \left( \frac{2a_k}{3} \right)^2 + 2 \cdot \frac{2a_k}{3} \cdot \frac{1}{a_k} + \left( \frac{1}{a_k} \right)^2.$

Simplify each term: $a_{k+1}^2 = \frac{4a_k^2}{9} + \frac{4}{3} + \frac{1}{a_k^2}.$

2. Using the induction hypothesis $a_k^2 > 3$, estimate the terms:
   • Since $a_k^2 > 3$, we know $\frac{4a_k^2}{9} > \frac{4 \cdot 3}{9} = \frac{4}{3}$.
   • Adding $\frac{4}{3}$ and $\frac{1}{a_k^2}$ (where $\frac{1}{a_k^2} > 0$), we conclude that: $a_{k+1}^2 > \frac{4}{3} + \frac{4}{3} = \frac{8}{3} > 3.$

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Step 4: Conclude by induction

By induction, $a_n^2 > 3$ for all $n \geq 0$.

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Summary

We have proven that $a_n^2 > 3$ for all $n$ using induction, relying on the sequence's recurrence relation and the assumption $a_k^2 > 3$ to establish $a_{k+1}^2 > 3$.

Would you like additional steps clarified or detailed further?

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Related Questions

1. What is the behavior of $a_n$ as $n \to \infty$?
2. Can we prove a stricter bound, such as $a_n^2 > 4$?
3. Is the sequence $a_n$ monotonic (increasing or decreasing)?
4. What happens if we change the initial condition, e.g., $a_0 = 1.5$?
5. How can we generalize this proof for similar recurrence relations?

Tip

When working with induction, always verify both the base case and the logical flow from $n = k$ to $n = k+1$.