What do you mean by:
"the inductive load should have low inrush current but it should have higher transient voltage."
You are still trying to infer that the inductor will have a larger effect on producing a dip in the supply when the inductor is initially activated.
Since the inductive load (such as the relay) has a resistive characteristics. As viewed from the output of the power supply, there is a sudden change of current (not inrush) because the load impedance decreases due to the low resistance of the coil. Note this is viewed from the output of the power supply, therefore, the load is the whole circuit. Let say initially, the circuit draws about 50mA (relay off), and rises up to 250mA when relay is activated. As seen by the power supply, there is a change of current for 200mA. We can say that it is caused by the relay.
With this change of current, if the power supply can not tolerate this change, there is a chance that the voltage supplied will be fluctuated. In order to resolve this, we can add a bulk capacitance (temporary energy storage) so that during the this sudden change of current, the voltage output is still stable and no fluctuation occur.
The problem by adding the capacitor is the inrush current (not change of current) during power up. As seen from the power supply, the capacitor is parallel with respect to the total load (the whole circuit). Therefore,
Ctotal=Cbulk+Ccircuit
The higher the capacitance, the higher the inrush current. This is because
I=C*dv/dt
So to resolve the inrush, we can add a resistor in series of the supply line. The trace impedance is not realiable to consider as your resistance. But if you design is part of a system (such as automotive) which would have long cables before the supply reaches your ciruit, this impedance is significant. This would render at least 1Ω of impedance and can reduce the inrush significantly.
As viewed from the relay coil, it can see a change of current from maybe 0A or a leakage of few µA to 200mA. With the characteristics of the inductor,
V=L*di/dt
it will have will rapid increase its voltage across itself and it will stop only when there is no more change of current (di). This phenomenon is called inductive kick off voltage. The problem here lies on the energy it gives off during the kick off (specially when you're using semicon in driving the relay). Since this energy is given, something should absord. If you design has no protection, it will destroy your relay driver or power supply or both. To resolve this, there are few solutions.
1. free-wheel diode that will absord the energy.
2. Transient suppressor such as transil, MOV, etc. that will clamp the kick off voltage so that the other circuits can tolerate the kick off voltage. But the suppressors partly absorb the energy too.
3. snubber, this is more on changing the waveform of the voltage ringing into lower frequency. From fast, high peak and destructive ringing into a more friendlier waveform where your components can tolerate.
These are all what I'm trying to infer here. I have simulated, calculated and design all of these situation. Maybe I stated something wrong but these are all I've seen during the past few years. If someone here who is knowledgable or an expert in design for EMC compatibility(specially the ISO pulses) please do correct me.