resistor combination circuit
R1 and R3 form a voltage divider.
This limits the voltage available to R2 to only 2.5 V.
The forward voltage of your LED is 2 V.
The remaining 0.5 V is dropped over R2, but this yields us a current of only:
$$\frac{0.5\ V}{4.7\ kΩ} =1\ mA$$
Which might not be enough to see it light.
Note that this is an over simplified way of calculating the voltages and currents (it's actually worse than I've shown).  I am ignoring certain effects, but the overall point is, you are drawing current over those resistances which is limiting the voltage and therefore current available to the upper LED.
If you could do away with R1, and increase R2 and R3 to compensate, this could work.