Let $\phi$ denote the Euler's phi function and $\sigma$ denote the sum of divisors function,then I want to show that $n^2>\phi(n)\sigma(n)$,I was trying to proceed using prime factorisation as follows:
Let $n=p_1^{\alpha_1}...p_k^{\alpha_k}$ then $\phi(n)\sigma(n)=\prod\limits_{i=1}^kp^{\alpha_i-1}(p^{\alpha_i+1}-1)$.
Now I observed that for $k=2$,
$n^2=p_1^{2\alpha_1}p_2^{2\alpha_2}>p_1^{2\alpha_1}p_2^{2\alpha_2}-p_1^{\alpha_1-1}p_2^{\alpha_2-1}.(p_1^{\alpha_1+1}+p_2^{\alpha_2+1}-1)=\phi(n)\sigma(n)$.
But for $k\geq 3$ there will be lots of product terms in $\phi(n)\sigma(n)$,then how to show the same inequality?
Continuing from what you've done,
$$\begin{equation}\begin{aligned} \prod_{i=1}^k p^{\alpha_i-1}(p^{\alpha_i+1}-1) & = \prod_{i=1}^k (p^{2\alpha_i}-p^{\alpha_i-1}) \\ & \lt \prod_{i=1}^k p^{2\alpha_i} \\ & = \left(\prod_{i=1}^k p^{\alpha_i}\right)^2 \\ & = n^2 \end{aligned}\end{equation}$$