1. ## Inequalities

X2+3+x-2>_7
is the above statement true for all real x>0?

Thanks

2. ## Re: Inequalities

What do you get when you expand $\left(x+\frac{1}{x}\right)^2$?

3. ## Re: Inequalities

Originally Posted by Trianagt
X2+3+x-2>_7
is the above statement true for all real x>0?

Thanks
\displaystyle \begin{align*} x^2 + 3 + x^{-2} &\geq 7 \\ x^4 + 3x^2 + 1 &\geq 7x^2 \\ x^4 - 4x^2 + 1 &\geq 0 \\ x^4 - 4x^2 + (-2)^2 - (-2)^2 + 1 &\geq 0 \\ \left(x^2 - 2\right)^2 - 3 &\geq 0 \\ \left(x^2 - 2\right)^2 &\geq 3 \\ \left|x^2 - 2\right| &\geq \sqrt{3} \\ x^2 - 2 \leq -\sqrt{3} \textrm{ or } x^2 - 2 &\geq \sqrt{3} \\ x^2 \leq 2 - \sqrt{3} \textrm{ or } x^2 &\geq \sqrt{2 + \sqrt{3}} \\ |x| \leq \sqrt{2 - \sqrt{3}} \textrm{ or } |x| &\geq \sqrt{2 + \sqrt{3}} \\ -\sqrt{2 - \sqrt{3}} \leq x \leq \sqrt{2 - \sqrt{3}} \textrm{ or } x \leq -\sqrt{2 + \sqrt{3}} \textrm{ or } x &\geq \sqrt{2 + \sqrt{3}} \end{align*}

4. ## Re: Inequalities

Originally Posted by Prove It
\displaystyle \begin{align*} x^2 + 3 + x^{-2} &\geq 7 \\ x^4 + 3x^2 + 1 &\geq 7x^2 \\ x^4 - 4x^2 + 1 &\geq 0 \\ x^4 - 4x^2 + (-2)^2 - (-2)^2 + 1 &\geq 0 \\ \left(x^2 - 2\right)^2 - 3 &\geq 0 \\ \left(x^2 - 2\right)^2 &\geq 3 \\ \left|x^2 - 2\right| &\geq \sqrt{3} \\ x^2 - 2 \leq -\sqrt{3} \textrm{ or } x^2 - 2 &\geq \sqrt{3} \\ x^2 \leq 2 - \sqrt{3} \textrm{ or } x^2 &\geq \sqrt{2 + \sqrt{3}} \\ |x| \leq \sqrt{2 - \sqrt{3}} \textrm{ or } |x| &\geq \sqrt{2 + \sqrt{3}} \\ -\sqrt{2 - \sqrt{3}} \leq x \leq \sqrt{2 - \sqrt{3}} \textrm{ or } x \leq -\sqrt{2 + \sqrt{3}} \textrm{ or } x &\geq \sqrt{2 + \sqrt{3}} \end{align*}
Is that necessary?

$x^2+3+x^{-2}=(x+x^{-1})^2+1$

5. ## Re: Inequalities

Originally Posted by a tutor
Is that necessary?

$x^2+3+x^{-2}=(x+x^{-1})^2+1$
It would be if you actually wanted to solve for x...

6. ## Re: Inequalities

Originally Posted by Prove It
It would be if you actually wanted to solve for x...
Yes.

You're short of things to do today aren't you.