Let $n=p_{1}^{\alpha_1} \cdot p_{2}^{\alpha_2} \cdot \cdot \cdot p_m^{\alpha_m}$. We know that $\tau (n) = \tau (p_1^{\alpha_1}) \cdot \tau (p_2^{\alpha_2}) \cdot \cdot \cdot \tau (p_m^{\alpha_m})$ and $\sigma (n) = (1+p_1+…+p_1^{\alpha_1})(1+p_2+…p_2^{\alpha_2}) \cdots (1+p_n+…+p_m^{\alpha_m})$. We will prove for most of the positive integers $n \sqrt{ \tau (n) } > \sigma(n)$. $$n \sqrt{ \tau (n) } > \sigma(n)$$$$p_{1}^{2\alpha_1} \cdot p_{2}^{2\alpha_2} \cdots p_m^{2\alpha_m} (\alpha_1 +1) \cdots (\alpha_m +1) > (1+p_1+…+p_1^{\alpha_1})(1+p_2+…p_2^{\alpha_2}) \cdots (1+p_n+…+p_m^{\alpha_m})$$Now look at the expression $$p_i^{2 \alpha_i} (\alpha_i +1) > (1+p_i+...+p_i^{\alpha_i})^2 = ( \frac{p_i^{\alpha_i +1}-1}{p_i-1} )^2$$$$p_i^{\alpha_i +2} \alpha_i + p_i^{\alpha_i} \alpha_i + p_i^{\alpha_i} + 2p_i > 2p_i^{\alpha_i +1} \alpha_i + 2p_i^{\alpha_i+1}$$For $p \ge 5$ we have the RHS bigger. For $p =2, 3$ we have solutions. If we check cases. Then we have $n=1, 2, 4, 6, 12$