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If $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ is the prime factorization of $n>1$ then show that :

$$1>\frac{n}{ \sigma (n)} > \left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\cdots\left(1-\frac{1}{p_r}\right)$$

I have solved the $1^\text{st}$ inequality($1>\frac{n}{ \sigma (n)}$) and tried some manipulations on the right hand side of the $2^\text{nd}$ inequality but can't get much further.Please help.

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    $\begingroup$ Did you mean $1-\frac1{p_2}$ instead of $1-\frac1{p_2}$? BTW if both sides are multiplicative functions, it suffices to prove the inequality for $n=p^a$, i.e. for powers of primes. $\endgroup$ – Martin Sleziak Jun 12 '12 at 17:41
  • $\begingroup$ @Chris oh sorry I have corrected $\endgroup$ – Saurabh Jun 12 '12 at 17:41
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Note that the function $\dfrac{n}{\sigma(n)}$ is multiplicative. Hence, if $n = p_1^{k_1}p_2^{k_2} \ldots p_m^{k_m}$, then we have that $$\dfrac{n}{\sigma(n)} = \dfrac{p_1^{k_1}}{\sigma \left(p_1^{k_1} \right)} \dfrac{p_2^{k_2}}{\sigma \left(p_2^{k_2} \right)} \ldots \dfrac{p_m^{k_m}}{\sigma \left(p_m^{k_m} \right)}$$ Hence, it suffices to prove it for $n = p^k$ where $p$ is a prime and $k \in \mathbb{Z}^+$.

Let $n=p^k$, then $\sigma(n) = p^{k+1}-1$. This gives us that $$\dfrac{n}{\sigma(n)} = p^k \times \dfrac{p-1}{p^{k+1}-1} = \dfrac{p^{k+1} - p^k}{p^{k+1} - 1} = 1 - \dfrac{p^k-1}{p^{k+1}-1}.$$ Since $p > 1$, we have that $p(p^k-1) < p^{k+1} - 1 \implies \dfrac{p^k-1}{p^{k+1}-1} < \dfrac1p \implies 1 - \dfrac{p^k-1}{p^{k+1}-1} > 1 - \dfrac1p$. Hence, if $n=p^k$, then $$\dfrac{n}{\sigma(n)} > \left( 1 - \dfrac1p\right)$$ Since, $\dfrac{n}{\sigma(n)}$ is multiplicative, we have the desired result.

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Let $S$ be the set of products of powers of the prime divisors of $n$ and $n/S$ the set of ratios $(n/s) : s \in S$.

$\frac{n}{\Pi (1 - \frac {1}{p_i})} > \sigma(n) \quad $ is the statement that the sum of all rational numbers in $n/S$ is larger than the sum of all integers in the same set.

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