Demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ for the given conditions. I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.
The proof provides me with the information that $x^2\gt 2$ and $x>0$.
I've taken the following steps to simplify the algebra...to the point where I arrive at:
$\frac{(x^2+2)^2}{(2*x)^2} = \frac {x^4+4*x^2+4}{4*x^2}$
I simplified this further to produce: $1 + \frac {1}{4}*x^2 + \frac{1}{x^2}$
In looking at this simplification, I see that I need to demonstrate (remembering the conditions), $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
Now, $\frac{1}{4}*x^2 \gt \frac{1}{2}$ and $\frac{1}{x^2}\lt \frac{1}{2}$ but I cannot figure out how to conclude that the addition of the two outputs something greater than 1. I imagine that I somehow need to show that the first term $\frac{1}{4}*x^2$ is somehow bigger than $0.5$ than $\frac{1}{x^2}$ is less than $0.5$.
In attempt to address this, I performed the following operation:
$\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} = \frac{1}{4}*x^4$
I know that $x^4$ must be greater than $4$...therefore $\frac{\frac{1}{4}*x^2}{\frac{1}{x^2}} \gt 1$.
Have I therefore successfully demonstrated that the addition of the two fractions:
$\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$
and can conclude that:
$(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$    ?
Edit: While the answers below also provide the solution to my question (albeit using a different strategy from the very beginning), I have figured out how to solve my question by finishing the procedure that I pursued above
Specifically, revisiting $\frac {1}{4}*x^2 + \frac{1}{x^2} \gt 1$, let's subtract $1$ from either side and then multiply through by $4x^2$ to produce:
$x^4-4x^2+4 \gt 0$  (multiplying by $4x^2$ will not change the direction of the inequality because $x\gt0$ )
We can factor this to $(x^2-2)^2$ and because $x \gt 0$, this number will always be greater than $0$.

It is equivalent to $$\frac{(x^2-2)^2}{4x^2}>0$$
