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If $p,q,r$ are real numbers and $0<p<q<r$, then $$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$

Is this a well-known inequality?

My proof of it is based on analytic geometry:

If you plot the points $P=(p,1/p)$, $Q=(q,1/q)$, $R=(r,1/r)$ in this order, you get the vertices of a triangle circumscribed by an equilateral hyperbola $xy=1$ entirely on the first quadrant (as all numbers $p,q,r$ are positive and because a line cannot intercept a conic in three points). The area of this triangle PQR can be given by the following formula: $$frac 12 begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix$$

And as these coordinates are written anticlockwise in this determinant, it has to be positive:
$$begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix>0$$

From that we get

$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$

Is this proof correct?

Is there another, simpler proof for this inequality?

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 $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
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.

Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ 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.

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 $$0<p<q<r$$