Is $f(x)=\left(\frac{x^2+1}{x^2}\right)^{x^2-x}$ an increasing function for $x \ge1$? The answer seems to me to be yes.  Here's my reasoning:
$\frac{x^2+1}{x^2}$ is a decreasing function since for $a > 0, x \ge 1$, $\frac{(x+a)^2+1}{(x+a)^2} < \frac{x^2+1}{x^2}$ since:

$$x^2[(x+a)^2+1]=x^2(x+a)^2+ x^2 < (x^2+1)(x+a)^2 = x^2(x+a)^2 + x^2+2ax+x $$ 

But for all $x$, $\frac{x^2+1}{x^2} > 1$ even while $\left(\frac{x^2+1}{x^2} - 1\right)$ is a decreasing function.
$(x^2-x)$ is clearly an increasing functions for $x \ge 1$ and therefore, so is $x^{(x^2-x)}$.
And for $a > 0, x\ge 1$, $\left(\frac{(x+a)^2+1}{(x+a)^2}\right)^{(x+a)^2-(x+a)} > \left(\frac{x^2+1}{x}\right)^{(x^2-x)}$ since:

$$\left(\frac{(x+a)^2-(x+a)}{x^2 -x}\right)\log\left(\frac{(x+a)^2+1}{(x+a)^2}\right) > \log\left(\frac{x^2+1}{x^2}\right)$$

The last step seems to be true but I don't know how to prove the inequality is correct.
Is my reasoning correct?  If so, how do I prove the last step?  Is there a simpler argument?
If my reasoning is not correct, what would be a simpler and more correct way to analyze whether $f(x) = \left(\frac{x^2+1}{x^2}\right)^{(x^2-x)}$ is increasing for $x \ge 1$?  

I think it's an increasing function.
Let $g(x)=(x^2-x)\ln\left(1+\frac{1}{x^2}\right)$.
Thus, $$g'(x)=(2x-1)\ln\left(1+\frac{1}{x^2}\right)-\frac{(x^2-x)\cdot\frac{2}{x^3}}{1+\frac{1}{x^2}}=(2x-1)\ln\left(1+\frac{1}{x^2}\right)-\frac{2(x-1)}{x^2+1}.$$
We'll prove that $g'(x)>0$ for all $x\geq1$.
Indeed, let $h(x)=\ln\left(1+\frac{1}{x^2}\right)-\frac{2(x-1)}{(2x-1)(x^2+1)}$.
Hence, $$h'(x)=-\frac{2(3x^3+3x^2-3x+1)}{x(2x-1)^2(x^2+1)^2}<0,$$
which says that $h$ is a decreasing function.
Thus, $h(x)\geq\lim\limits_{x\rightarrow+\infty}h(x)=0$ and we are done!
