We only give a sketch of parts of the proof, leaving all the details
to the reader.
Set 
so that 
. On the other hand, set
An explicit computation shows that we have
We express these quantities on purpose only in terms of b, P , Q and D,
and will use the inequalities 
.
When we want to find an upper bound for an expression in a, b, c, d,
we consider this expression first as a function of the single parameter b
which is constrained to satisfy 
. This is done by taking
derivatives and looking at the value of the expression where the derivative
vanishes and at the endpoints. We then have a function of P, Q and D
only. In most cases, this expression does not even depend on P, and hence
we again compute derivatives and look at endpoints to find the maximum as
Q varies from 0 to 
. When the expression depends on P, we use
the inequalities 
.
We give one example, the bound for bc. We set 
. Then
We find that the derivative in B of the
RHS vanishes for 
and for these values
we have 
, the 
signs
being independent. The study of the 4 functions
on the interval 
shows that the largest maximum is attained with both
+ signs for x=1, and is equal to 12, hence a local maximum of |bc|
is attained for 
and is equal to 
. Furthermore, looking at
the endpoints B=0 and 
, we see that bc is respectively
0 and 
, which is smaller that 
thus
giving the desired inequality for |bc|. However, in addition we have
, hence since a>0, 
if and only if 
.
When 
, we must have 
, hence in the expression for a the
expression under the square root must be larger than 
, or equivalently
which shows the lower inequality for b. Hence 
implies c<0. This implies that if bc<0, we must have b>0 and c<0.
But in that case, the - sign must hold in the expression for c, and hence
since b>0, in the 4 functions above the second 
sign must in fact
be a - sign. If we look at the remaining two functions, we see that the
local maximum of |bc| is now only 
. The endpoints are
B=0 and B=P in this case, but then bc vanishes at both endpoints so
the maximum is really 
for bc<0.
For the two inequalities involving 
, we get in fact a fourth degree
equation in 
, but luckily this equation has B=P as double root, hence
we are reduced to a second degree equation one again.
The details of this and of the other inequalities are left to the patient
reader.
