How Does the Condition x > -1 Influence the Inequality x² + 1/(x²+1) ≥ 1?

AI Thread Summary
The discussion centers on proving that if x > -1, then x² + 1/(x² + 1) ≥ 1. Participants note that x > -1 does not necessarily imply x² > 1, as counterexamples exist. They suggest analyzing two cases based on the absolute value of x: one where |x| ≥ 1 and another where |x| < 1. The importance of the condition x > -1 is debated, with some questioning its necessity for establishing the inequality. Ultimately, the conversation highlights the need for a detailed case analysis to arrive at the conclusion.
Nipuna Weerasekara
Messages
36
Reaction score
2

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...
 
Physics news on Phys.org
Nipuna Weerasekara said:

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##.
 
Nipuna Weerasekara said:
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...
 
Nipuna Weerasekara said:
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##.

Deleted
 
Nipuna Weerasekara said:

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?
 
Back
Top