# Conditional Proof question

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1. Nov 7, 2016

### Nipuna Weerasekara

1. The problem statement, all variables and given/known data
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...

2. Nov 7, 2016

### Staff: Mentor

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. $x = -\frac{1}{2}$.

3. Nov 7, 2016

### Nipuna Weerasekara

Yup you're correct, but then how do I do it. I have no idea how to get something like $x^2\geq1$ from $x>-1$.

4. Nov 7, 2016

### Staff: Mentor

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.

5. Nov 7, 2016

Thank you...

6. Nov 7, 2016

Deleted

7. Nov 7, 2016

### Ray Vickson

Is the restriction $x >-1$ important for getting the inequality?