Getting output of 9v DC from 5v DC?
Those relays say 12v on them, not 9v. 

Even if they do pull in at 9v, you're on the edge. An increase in temperature would increase their resistance and therefore lower the current.

To take up replete's suggestion, this is how you might attempt to provide a higher voltage to pull the relays in, and then sustain with a lower voltage. This is usually possible as a closed relay has a better magnetic path than an open relay, so the drop out current is much lower than the pull in current.







simulate this circuit &ndash; Schematic created using CircuitLab

R1 and R2 charge C1 to full rail potential while Q1 is off. They need to be small enough to do this in the time you have available, and large enough not to lose too much charge during the pull in sequence.

When you want the relay to pull in, drive both transistors on. Q1 pulls the bottom of C1 up to rail, meaning nearly 2x rail appears through D1 at the relay. C1 must be large enough to power the relay coil during pull in, while not dropping to below the pull in voltage. 

Once C1 has discharged, the relay is sustained by nearly the rail voltage through D2. These would be schottky diodes for minimum voltage drop. Q1 can be turned off at this point, or left on. C1 doesn't recharge until Q1 is turned off.

To release the relay, turn off Q2.