Probably the simplest demonstration in the world that $\;\frac{a}{b}+\frac{b}{a}\geqslant 2$

           We will prove that

$$\frac{a}{b}+\frac{b}{a}\geqslant 2,\;\forall a,b>0.$$

     If $\;a=b$ then $\frac{a}{b}+\frac{b}{a}=2.$

     If $\;0<a<b$ then $\frac{b-a}{a}>\frac{b-a}{b}\;\tag{1}$

this is because the fractions have the same numerator, and the denominator of the first fraction is smaller than the denominator of the second fraction.

But inequality (1) is written equivalently $\frac{b}{a}-1>1-\frac{a}{b}\;\Leftrightarrow\;\frac{a}{b}+\frac{b}{a}>2.$

          In conclusion $\frac{a}{b}+\frac{b}{a}\geqslant 2\;,\;\forall a,b>0\;$ with equality if and only if $a=b$.

$\blacksquare$