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Problem:

Let $x$ and $y$ be positive real numbers. Let $a_0 =y$. Show that $$a_n = \frac12\left(\frac{x}{a_{n-1}} + a_{n-1}\right)$$ is a decreasing sequence.

Solution Attempt:

I have tried showing that the difference is positive $$a_{n-1} - a_{n} > 0,$$ that the ratio is greater than 1 $$\frac{a_n}{a_{n-1}} < 1,$$ and I have tried to use induction by considering $a_0$ and $a_1$. Somewhere, I must be doing something wrong. I keep getting the result $a_n > \frac{1}{2}a_{n-1},$ which doesn't seem very useful.

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    i think induction will help you – 2017-02-20
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    When $a_0 = y = 0.1, x = 2$ we get $a_1 = 10.05$, which is contradicting to the claim right? – 2017-02-20
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    @stud_iisc You're right. We need $y >\sqrt{x}$. See my answer. – 2017-02-20

2 Answers 2

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Your claim is not true. We need a condition that $a_{0}=y \ge \sqrt{x}$.

We can use that $$a_{n}=\frac{1}{2} \left(a_{n-1}+\frac{x}{a_{n-1}} \right) \ge \sqrt{x}$$ By $\text{AM-GM}$. So $a_{n} \ge \frac{x}{a_{n}} $ for all $n$. Now note that $$a_{n+1}= \frac{1}{2} \left(a_{n}+\frac{x}{a_{n}} \right) \le \frac{1}{2} (a_{n}+a_{n})=a_{n} $$

From $(1)$. So $a_{n+1} \le a_{n}$. The sequence is monotonically decreasing.

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    What does AM-GM stand for? – 2017-02-20
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    @EternusVia I added a link. – 2017-02-20
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    It seems like your proof doesn't need the condition $y \geq \sqrt{x}$. Where does this assumption come into play? Is it a sort of implicit induction? – 2017-02-20
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    @EternusVia We need to prove that that $a_{0}>a_{1}$. – 2017-02-20
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    I get it now. Thanks! – 2017-02-20
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    Why have you replaced $x$ with $n$? $a_n\ne \frac12\left(a_{n-1}+\frac{n}{a_{n-1}}\right)$. – 2017-02-20
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    @Dr.MV Oh shoot. Good point. – 2017-02-20
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    @Dr.MV Fixed :) – 2017-02-20
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Your claim is NOT true in general. It is only true if $y\ge\sqrt{x}$.

However, if $y<\sqrt{x}$, then one can show the following: $$ a_0<\sqrt{x}

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    How were you able to determine these conditions on $y$? – 2017-02-20
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    This seems untrue. For example, if $a_{0}=1$, $x=2$, then we have that $a_{1}=\frac{3}{2}$, $a_{2}=\frac{17}{12}>\sqrt{2}$. So you're incorrect. – 2017-02-20
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    I have corrected my answer accordingly. Thanx. – 2017-02-20