# Thread: Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

1. ## Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

Thanks a lot!

The whole problem is:

"Let $f(x)=(x^2+3)/(2x)$ for $x \not= 0$. Define a sequence of real numbers $x_n$ by

$x_{n+1} =f(x_n)$ for $n \geq 1$, $x_1=2$.

A) Show that if $x> \sqrt 3$, then $f(x)> \sqrt 3$.
B) Show that if $x> \sqrt 3$, then $x>f(x)$. (Hint: show $x-f(x)>0$.)
C) Conclude that the sequence $x_n$ converges."

2. ## Re: Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

Originally Posted by buenogilabert
Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

Thanks a lot!

The whole problem is:

"Let $f(x)=(x^2+3)/(2x)$ for $x \not= 0$. Define a sequence of real numbers $x_n$ by

$x_{n+1} =f(x_n)$ for $n \geq 1$, $x_1=2$.

A) Show that if $x> \sqrt 3$, then $f(x)> \sqrt 3$.
B) Show that if $x> \sqrt 3$, then $x>f(x)$. (Hint: show $x-f(x)>0$.)
C) Conclude that the sequence $x_n$ converges."
A) $f(x)=\frac{x^2+3}{2x}$

$=\frac{x}{2}+\frac{3}{2x}$

$\ge 2\sqrt{\frac{x}{2}\cdot \frac{3}{2x}}$ (AM-GM inequality, which holds for $x>0$)

$=\sqrt{3}$

So, if $x>0$, then $f(x)\ge \sqrt{3}$.

Equality holds only when

$\frac{x^2+3}{2x}=\sqrt{3}$

$x^2+3=2\sqrt{3}x$

$x^2-2\sqrt{3}x+3=0$

$(x-\sqrt{3})^2=0$

$x=\sqrt{3}$

So, if $x>0$ and $x\neq\sqrt{3}$, then $f(x)>\sqrt{3}$.

In particular, if $x>\sqrt{3}$, then $f(x)>\sqrt{3}$.