Prove the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$ satisfies $a_{n} > 2$

Question

Prove the sequence $a_{1} = 4$, $a_{n + 1} = \frac{a_{n}}{2} + \frac{2}{a_{n}}$, $n = 1, 2, \ldots$ satisfies $a_{n} > 2$

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Let $x = a_{n}/2$.

Then $a_{n + 1} = x + 1/x$.

Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 \implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.

This shows $a_{n + 1} \geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.

Answer 1

AM-GM gives: $$\frac{a}{2} + \frac{2}{a} \stackrel{a>0, a\neq 2}{>} 2\sqrt{\frac{a}{2}\cdot\frac{2}{a}} = 2$$

Edit after comments:

• Note that $a^{\star} = 2$ is a fixpoint of the iteration as $\frac{a^{\star}}{2} + \frac{2}{a^{\star}} = 2$.
• For any other starting value $a_0 > 0, a_0 \neq 2$ AM-GM tells you, that the iteration $a_{n+1} = \frac{a_n}{2} + \frac{2}{a_n}$ produces $a_{n+1} > 2$.

So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.

For a direct proof that all members of the iteration are greater than $2$ you may consider

• $f(x) = \frac{x}{2} + \frac{2}{x} \Rightarrow f'(x) = \frac{1}{2} - \frac{2}{x^2}$

Now use MVT:

$$f(x) - 2 = f(x) - f(2) = f'(\xi)(x - 2) = \left(\frac{1}{2} - \frac{2}{{\xi}^2}\right) (x - 2) \stackrel{2 < \xi < x}{>} 0$$

Answer 2

Use principle of mathematical induction,

Basis for induction $$a_1>2$$

Induction Hypothesis $$a_k>2$$ $$\Longleftrightarrow \frac{a_k}{2}>1$$

Inductive step $$a_{k+1}=\frac{a_k}{2}+\frac{2}{a_k}>2$$ which is true by condition of equality in AM-GM inequality and induction hypothesis.

Hence proved

Hope it is helpful

Answer 3

Following with your reasoning, suppose $x + 1/x = 2$. Then: $$x + 1/x = 2\implies x^2 - 2x + 1 = 0\implies x = 1.$$ But this is impossible because...

Answer 4

Note $a_n >0$, $n=1,2,3,.....$.

Let $n \ge 1$:

$a_{n+1}= \dfrac{a_n^2 +4}{2a_n}=$

$\dfrac{(a_n-2)^2+4a_n}{2a_n}=$

$\dfrac{(a_n-2)^2}{2a_n} + 2 \ge 2$.

(The first term $\ge 0$, a square divided by a positive number.)

Hence

$a_{n+1} \ge 2$, $n \in \mathbb{N}$.