Proof of an inequality using analytic geometry

Put $x={r\over q}>1$ and $y= {q\over p}>1$ so we have to prove $${1\over x}+{1\over y} +xy >x+y+{1\over xy}$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$ or $$x^2y^2-xy(x+y)+(x+y)-1>0$$

or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$

or $$(xy-1)(x-1)(y-1)>0$$ which is true.

Just a slightly different view at your proof: If the function $f: I \to \Bbb R$ is strictly convex on the interval $I \subset \Bbb R$ then for $p < q < r$ in $I$ $$ f(q) < \frac{r-q}{r-p} \, f(p) + \frac{q-p}{r-p} \, f(r) \\ \iff (r-q) \, f(p) + (p-r) \, f(q) + (q-p) \, f(r) > 0 \quad (*) $$ Choosing $f(x) = \frac 1x$ gives the desired inequality.

The connection to your solution is that $(*)$ can be written as $$ \begin {vmatrix} p & f(p) & 1 \\ q & f(q) & 1 \\ r & f(r) & 1 \\ \end {vmatrix}>0 \, , $$ i.e. the (oriented) area of the triangle is positive because $f(x) = \frac 1x$ is strictly convex.

Your inequalitiy is equavalent to $$- \left( q-r \right) \left( p-r \right) \left( p-q \right) >0$$ and this is true since we have $$0<p<q<r$$