Proving the formula for capacitance multiplier
Your diagram was too complicated, so I've redrawn it:



simulate this circuit &ndash; Schematic created using CircuitLab
Note that this now highlights the fact that \$V_i\$ is a voltage source. And rather than &quot;busing around&quot; that source I've separated it out. I also just tossed down a couple of node labels.
Here, we can say that there is a nice voltage divider in \$R_2\$ and \$C_1\$:
$$V_a=V_i\cdot \frac{Z_{C_1}}{R_2+Z_{C_1}}=V_i\cdot \frac{\frac1{s\,C_1}}{R_2+\frac1{s\,C_1}}=V_i\cdot \frac1{1+R_2\,C_1\,s}$$
But on the ideal opamp assumption for this follower arrangement, we can also say that \$V_b=V_a\$.
Also, we know \$I_{R_1}=\frac{V_i-V_b}{R_1}\$ and \$I_{R_2}=\frac{V_i-V_a}{R_2}\$. (Of course, we can replace \$V_b\$ with \$V_a\$ for \$I_{R_1}\$.)
To work out the impedance that \$V_i\$ sees, we just need to calculate:
$$\begin{align*}
Z&amp;=\frac{V_i}{I_{R_1}+I_{R_2}}\\\\
&amp;=\frac{V_i}{\frac{V_i-V_a}{R_1}+\frac{V_i-V_a}{R_2}}\\\\
&amp;=\frac{V_i}{\frac{V_i-V_i\cdot \frac1{1+R_2\,C_1\,s}}{R_1}+\frac{V_i-V_i\cdot \frac1{1+R_2\,C_1\,s}}{R_2}}\\\\
&amp;=\frac{1}{\frac{1-\frac1{1+R_2\,C_1\,s}}{R_1}+\frac{1-\frac1{1+R_2\,C_1\,s}}{R_2}}\\\\
&amp;=\frac{R_1\left(1+R_2\,C_1\,s\right)}{\left(R_1+R_2\right)\,C_1\,s}\\\\
&amp;=\frac{1+R_2\,C_1\,s}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}\\\\
&amp;=\frac{1}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}+\frac{R_2}{1+\frac{R_2}{R_1}}\\\\
&amp;=\frac{1}{\left(1+\frac{R_2}{R_1}\right)\,C_1\,s}+\left(R_1\mid\mid R_2\right)
\end{align*}$$
At this point, it's pretty clear that the unitless \$1+\frac{R_2}{R_1}\$ multiplies \$C_1\$ and that the first term is the impedance of this multiplied capacitance. But because of the sum, it's in series with the second term which is the parallel resistance of \$R_1\$ and \$R_2\$. That makes sense as both resistors have \$V_i\$ on one side and \$V_a\$ (virtual or otherwise) on the other side.
So the two resistors really are effectively in parallel with each other, but in series with the multiplied capacitance.
None of this qualifies as an &quot;intuitive approach.&quot; (Unless one considers pursuing a chain of algebraic manipulation as intuitive.) But you said you were looking for &quot;the mathematical justification&quot; and that you &quot;have to prove that [...] by means of mathematics.&quot; So that's the direction I went here.