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

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.

All these and a lot more are covered here, including linked questions ... — rtybase, Dec 12 '18 at 09:22

trancelocation · Answer 1 · 2018-12-12T09:35:50.407

4

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:

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

edited Dec 12 '18 at 09:35

answered Dec 12 '18 at 05:49

trancelocation

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how do i know $a \neq 2$? – Dec 12 '18 at 06:31
@stackofhay42: If $a=2$, we t get the AM-GM $\textbf{equality}$. – Yadati Kiran Dec 12 '18 at 06:47
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning. – Dec 12 '18 at 06:51
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,\Longrightarrow a_{k+1}>0$ by the given relation. – Martund Dec 12 '18 at 09:25
@stackofhay42: We can also prove that by induction. $a_1\neq 2, a_k\neq 2 \Longrightarrow a_{k+1}\neq 2$ by the given relation. – Martund Dec 12 '18 at 09:27

score 3 · Answer 2 · answered Dec 12 '18 at 06:08

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

score 2 · Answer 3 · answered Dec 12 '18 at 11:01

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

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

Martín-Blas Pérez Pinilla · Answer 4 · 2018-12-12T10:44:04.710

0

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...

edited Dec 12 '18 at 10:44

answered Dec 12 '18 at 07:44

Martín-Blas Pérez Pinilla

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What about $a_n<2$? – Rakesh Bhatt Dec 12 '18 at 10:14
@RakeshBhatt, the OP has proved $a_n > 2\implies a_{n+1}\ge 2$. Only is required excluding the equality. – Martín-Blas Pérez Pinilla Dec 12 '18 at 10:43
Why is this impossible? – Ekesh Kumar Dec 13 '18 at 23:45
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$). – Martín-Blas Pérez Pinilla Dec 14 '18 at 07:12

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

4 Answers4