Let be a reduced form such that . We have the following inequalities:

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 haveWe express these quantities on purpose only in terms of

b,P,QandD, and will use the inequalities . When we want to find an upper bound for an expression ina,b,c,d, we consider this expression first as a function of the single parameterbwhich 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 ofP,QandDonly. In most cases, this expression does not even depend onP, and hence we again compute derivatives and look at endpoints to find the maximum asQvaries from0to . When the expression depends onP, we use the inequalities .We give one example, the bound for

We find that the derivative inbc. We set . ThenBof 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 forx=1, and is equal to12, hence a local maximum of|bc|is attained for and is equal to . Furthermore, looking at the endpointsB=0and , we see thatbcis respectively0and , which is smaller that thus giving the desired inequality for|bc|. However, in addition we have , hence sincea>0, if and only if .When , we must have , hence in the expression for

athe expression under the square root must be larger than , or equivalently which shows the lower inequality forb. Hence impliesc<0. This implies that ifbc<0, we must haveb>0andc<0. But in that case, the-sign must hold in the expression forc, and hence sinceb>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 areB=0andB=Pin this case, but thenbcvanishes at both endpoints so the maximum is really forbc<0.For the two inequalities involving , we get in fact a fourth degree equation in , but luckily this equation has

B=Pas 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.