is there any analytical way to konw if $\frac{1}{2x}+\frac{x}{2} >1$ for $(1,\infty)$ or $(0,\infty)$? I was solving a physics problem and I got
$$\frac{a^2+b^2}{2ab}$$
or equivalents
$$\frac{1+ \left( \frac{b}{a}\right)^2}{2\cdot\frac{b}{a}}.$$
If $x=\frac{b}{a}$, we obtain 
$$\frac{1+x^2}{2x}$$ or $$\frac{1}{2x}+\frac{x}{2}.$$
I need to know if this expresions are $>1$ when $a<b$ (or $x>1$).
I know the answer is yes (I used geogebra to plot) but I'd like to know if there is any analytical way to get the answer. Thank you.
 A: From your expression $\frac{x}{2}+\frac{1}{2x}$ for $x>0$ you can use AM-GM:
$$\frac{x}{2}+\frac{1}{2x}\geq2\sqrt{\frac{x}{2}\cdot\frac{1}{2x}}=1.$$
The equality occurs for $x=1$, which says that $1$ is a minimal value.
If $x$ real number without condition $x>0$ then since $\lim\limits_{x\rightarrow0^-}\left(\frac{x}{2}+\frac{1}{2x}\right)=-\infty$,
we see that the minimum of this expression does not exist.
A: Note that
$$0\leq (a-b)^2 = a^2 - 2ab + b^2$$
and hence
$$a^2+b^2 \geq 2ab.$$
Therefore, assuming $ab>0$, we have
$$ \frac{a^2+b^2}{2ab} \geq 1.$$
A: For $x > 0$  (and $x$ must be greater than zero):
$\frac {1}{2x} + \frac x2 > 1 \iff$
$\frac 1x + x > 2 \iff$
$1 + x^2 > 2x \iff $
$x^2 - 2x + 1 > 0 \iff$
$(x - 1)^2 > 0 \iff$
$x - 1 \ne 0 \iff$
$x \ne 1$.
That's it.  If $x = 1$ then $\frac 1{2x}+ \frac x2 = 1$ other wise $\frac 1{2x}+ \frac x2 > 1$
If $x < 0$ well .... $\frac 1{2x}  + \frac x2 < 0 < 1$ so duh... but we can use the same argument to show that $\frac 1{2x} +\frac x2 < -1$ for $x < 0; x\ne -1$
