Question 1208897
<pre>
In the above I used this without proof:

If {{{0<x<y<z}}} then {{{matrix(1,2,0<1/z<1/y<1/x,"")}}}

Proof:

{{{0<x<y<z}}}

Multiply through by x, then by y, then by z

{{{matrix(1,2,system(0<x^2<xy<xz,0<xy<y^2<yz,0<xz<yz<z^2),"")}}}

From the 1st and 3rd inequalities (we don't need the middle one),
we get

{{{0<xy<xz<yz}}}

Now we divide through by xyz

{{{0<(xy)/(xyz)<(xz)/(xyz)<(yz)/(xyz)}}}

{{{matrix(1,2,0<1/z<1/y<1/x,"")}}}

Edwin</pre>