Proof of an inequality using analytic geometry

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?

Solution 1:

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.

Solution 2:

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.