Here are two op-amp circuits. Both have a unity inverting signal gain.
A basic inverting amplifier

An inverting amplifier with 30k resistor connected between the op-amp's inverting input and 0V.

At first it may seem like the one with R3 aught to have a lower gain and it does by a small margin, especially at high frequencies, but the inverting input is a virtual ground node, so it's not that straightforward. There are two effects which cancel one another out.
R2 and R3 form a potential divider, which should divide the input signal by 5. V
OUT = V
IN*R3/(R2+R3) and when V
IN = 1V:
V
OUT = 1*30k/(120k+30k) = 0.2V.
But the output impedance of the potential divider is equal to the value of the two resistors in parallel.
R
OUT = (120k*30k)/(120k+30k) = 24k
This can be simplified to an equivalent circuit, showing 0.2V in series with 24k.
When connected to the op-amp circuit, the gain is now 5.

So we have an attenuator dividing the signal by a factor of 5, connected to an op-amp with an inverting gain of 5, giving a total gain of 1.
Now, imagine the input is connected to zero and a noise voltage is injected into the inputs. To simplify matters assume it's the non-inverting input. Say the noise voltage is 10nV. The circuit is now a non-inverting amplifier. Remove R3 and it has a gain of Av = 1+ 120k/120k = 2, keep it and you have a gain of 6. So assuming there are no other noise sources, the output noise will be 20nV, without R3 and 60nV with it in place.

So why might an op-amp only be stable when the noise gain is over a certain figure?
The op-amp has phase shift and if there's still gain, when the phase shift is 180º it will oscillate, since negative feedback, will become positive. The resistors in the feedback network, attenuate the op-amp's output, before it's fed back into the negative input. If the op-amp has a gain of 5, when the phase shift is 180º, then it will oscillate if the resistive divider attenuates the signal by less than a factor of 5. In the circuit with the noise gain of 6, but signal gain of unity, the resistive divider is attenuating the output by a factor of 6, so it won't oscillate, if the op-amp has a gain of 5, with a 180º phase shift.
In practice, the circuit can still ring, even if the gain us under unity, when the phase shift is 180º, as it will be under-damped, so op-amps are designed to have a phase margin or less than that, so they're over-damped.
To reduce the noise at lower frequencies, yet still avoid oscillation, R3 can be connected to 0V, via a capacitor. Here's a demonstration of a noise analysis. I've made the input noise of the op-amp stupidly high, at 10mV, (en = 10mV) so it dominates. At low frequencies the noise gain is 2, at higher frequencies it's 6, followed by a roll-off at much higher frequencies, due to the op-amp running out of gain.
