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Conditional Proof question

1. Homework Statement
Let ##x\in \mathbb{R} ##
Prove the conditional statement that,
if ## x>-1## then ## x^2 + \frac {1}{x^2+1} \geq 1##

2. The attempt at a solution

Suppose ## x>-1## is true.
Then ## x^2>1##
Then ## \frac{1}{2}>\frac {1}{x^2+1}##
Then ##x^2+ \frac{1}{2}>x^2+\frac {1}{x^2+1}##

After that I have no clue how to get to the part where ## x^2 + \frac {1}{x^2+1} \geq 1## happens. Pls help...
 

fresh_42

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1. Homework Statement
Let ##x\in \mathbb{R} ##
Prove the conditional statement that,
if ## x>-1## then ## x^2 + \frac {1}{x^2+1} \geq 1##

2. The attempt at a solution

Suppose ## x>-1## is true.
Then ## x^2>1##
Then ## \frac{1}{2}>\frac {1}{x^2+1}##
Then ##x^2+ \frac{1}{2}>x^2+\frac {1}{x^2+1}##

After that I have no clue how to get to the part where ## x^2 + \frac {1}{x^2+1} \geq 1## happens. Pls help...
Aren't you already done if ##x^2 \geq 1\;##? Isn't ##\frac{1}{1+x^2}## simply something positive?
But ##x > -1## doesn't imply ##x^2 > 1##, e.g. ##x = -\frac{1}{2}##.
 
Yup you're correct, but then how do I do it. I have no idea how to get something like ##x^2\geq1## from ## x>-1##.
 

fresh_42

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2018 Award
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Yup you're correct, but then how do I do it. I have no idea how to get something like ##x^2\geq1## from ## x>-1##.
You can distinguish between the two cases.
  1. ##|x| \geq 1##
  2. ##|x| < 1##
In the first case you have (together with ##x >-1##) that ##x \geq 1## and ##x^2 \geq 1##.
In the second case, have a look on ##\frac{1}{x^2+1}## first.
 
Thank you...
 

Math_QED

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Ray Vickson

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1. Homework Statement
Let ##x\in \mathbb{R} ##
Prove the conditional statement that,
if ## x>-1## then ## x^2 + \frac {1}{x^2+1} \geq 1##

2. The attempt at a solution

Suppose ## x>-1## is true.
Then ## x^2>1##
Then ## \frac{1}{2}>\frac {1}{x^2+1}##
Then ##x^2+ \frac{1}{2}>x^2+\frac {1}{x^2+1}##

After that I have no clue how to get to the part where ## x^2 + \frac {1}{x^2+1} \geq 1## happens. Pls help...
Is the restriction ##x >-1## important for getting the inequality?
 

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