Question 361025
Let {{{b = 1/t^2}}}.
Proof:

If{{{b = 1/t^2}}}, then because b < x, {{{1/t^2<x}}}.
{{{1<xt^2}}}, since t>0.
{{{0<t^2x-1 = (t*sqrt(x)-1)(t*sqrt(x)+1)}}}, after transposition and factoring the quadratic expression on the right.
{{{0<t*sqrt(x)-1}}}, after dividing both sides by {{{t*sqrt(x)+1}}}, which is surely positive.  So
{{{1<t*sqrt(x)}}},and thus
{{{1/sqrt(x)<t}}}, since {{{sqrt(x)}}}is positive.