# Show the inequality $\sigma(n)\phi(n) \geq n^2(1-\frac{1}{p_1^2})(1-\frac{1}{p_2^2})\cdots(1-\frac{1}{p_r^2})$

If $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$then ,show the inequality :$$\sigma(n) \phi(n) \geq n^2(1-\frac{1}{p_1^2})(1-\frac{1}{p_2^2})\cdots(1-\frac{1}{p_r^2})$$

I know the function $\sigma(n) \phi(n)$ is multiplicative hence I only have to show the inequality for $n=p^{k}$
So $$\sigma(n) \phi(n)$$ $$\implies \frac{n(p^{k+1}-1)}{p}$$ But after then how to get the desired inequality?

• $n\frac{p^{k + 1} - 1}{p} = n\left( p^k - \frac{1}{p} \right) = n\left( n - \frac{1}{p} \right) = n^2\left( 1 - \frac{1}{np} \right) \geq n^2\left( 1 - \frac{1}{p^2} \right)$ – TenaliRaman Jun 16 '12 at 14:33

## 1 Answer

Putting $n = p^k$, $\sigma(n)\phi(n) = \frac{p^{k + 1} - 1}{p - 1} p^{k -1}(p - 1) = p^{2k} - p^{k - 1}$ and $n^2 (1 - \frac{1}{p^2}) = p^{2k} - p^{2k - 2}$.

So we get the inequality and furthermore we see that equality happens iff k = 1. Thus for a general $n$ the equality happens iff $n$ is squarefree.