Author Topic: Inequality (Read 1879 times) Tweet Share

/0 · « **on:** April 23, 2009, 01:32:31 pm »

Prove $x+\frac{1}{x} \geq 2$ for $x > 0$ WITHOUT CALCULUS

kamil9876 · « **Reply #1 on:** April 23, 2009, 03:16:30 pm »

You can use the method i used for Ahmad's problem:
http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

checkout the problem in my second post of that thread, it has a similair strategy too

Your problem:

It's true for x=1. if increase x from 1 to 1+a:

then the expression: $x+\frac{1}{x}$ becomes:

$1+a + \frac{1}{1+a}$
$=2+a-\frac{a}{a+1}$ (1)

obviously -1<a since x is positive. If a was positive then $a>\frac{a}{a+1}$ [/tex] and so $a-\frac{a}{a+1}$
is positive which means that expression 1 is greater than two as required. However if -1<a<0 then let a=-b and so 0<b<1.

$a-\frac{a}{a+1}=\frac{b}{1-b}-b$

But obviously $\frac{b}{1-b}$ has a larger modulus than b. and so $a-\frac{a}{a+1}=\frac{b}{1-b}-b$ is positive in this case too.

/0 · « **Reply #2 on:** April 23, 2009, 07:31:50 pm »

nice! thanks kamil

Mao · « **Reply #3 on:** April 23, 2009, 09:25:09 pm »

$x + \frac{1}{x} \ge 2$

$\implies x^2 + 1 \ge 2x, \; x>0$

$\implies x^2 - 2x + 1 \ge 0,\; x>0$

$\implies (x-1)^2 \ge 0$ , which is true

$\therefore x + \frac{1}{x} \ge 2, \; x>0$

TrueTears · « **Reply #4 on:** April 23, 2009, 09:29:14 pm »

Interesting methods lol

TonyHem · « **Reply #5 on:** April 23, 2009, 09:31:27 pm »

Quote from: Mao on April 23, 2009, 09:25:09 pm

$x + \frac{1}{x} \ge 2$

$\implies x^2 + 1 \ge 2x, \; x>0$

$\implies x^2 - 2x + 1 \ge 0,\; x>0$

$\implies (x-1)^2 \ge 0$ , which is true

$\therefore x + \frac{1}{x} \ge 2, \; x>0$

omg I had that too!
haha

Search the forums now!

We have moved!

Author Topic: Inequality (Read 1879 times) Tweet Share

/0

Inequality

kamil9876

Re: Inequality

/0

Re: Inequality

Mao

Re: Inequality

TrueTears

Re: Inequality

TonyHem

Re: Inequality

Recent Posts

User:
Password: