is there any analytical way to konw if $\frac{1}{2x}+\frac{x}{2} >1$ for $(1,\infty)$ or $(0,\infty)$?

Question

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.

Answer 1

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 $$ \frac{a^2+b^2}{2ab} \geq 1.$$

Answer 2

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.

Answer 3

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 $x < 0; x\ne -1$