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Let $x \in \mathbb{N}$, and let $$x = \prod_{i=1}^{\omega(x)}{{p_i}^{\alpha_i}}$$ be the canonical factorization of $x$. (That is, the $p_i$'s are primes with $p_1 < \ldots < p_{\omega(x)}$.) Note that $\omega(x)$ is the number of distinct prime factors of $x$.

What is wrong with this proof that $p_1 > \omega(x)$, where $p_1$ is the least prime dividing $x$?

Let $\sigma(x)$ be the sum of the divisors of $x$. Then the abundancy index $I(x) = \sigma(x)/x$ is bounded above by $$I(x) = I\left(\prod_{i=1}^{\omega(x)}{{p_i}^{\alpha_i}}\right) = \prod_{i=1}^{\omega(x)}{I\left({p_i}^{\alpha_i}\right)} < \prod_{i=1}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)}.$$

Now, I use the inequality $$\frac{p_i}{p_i - 1} < \frac{p_1 - (i - 1)}{p_1 - i}$$ for $i \neq 1$. This yields $$1 \leq I(x) < \prod_{i=1}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)} = \left(\frac{p_1}{p_1 - 1}\right)\prod_{i=2}^{\omega(x)}{\left(\frac{p_i}{p_i - 1}\right)} < \left(\frac{p_1}{p_1 - 1}\right)\prod_{i=2}^{\omega(x)}{\left(\frac{p_1 - (i - 1)}{p_1 - i}\right)} = \frac{p_1}{p_1 - \omega(x)}.$$

Finally, I have $p_1 > \omega(x)$.

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    $\begingroup$ Take $x=30..........$ $\endgroup$ – Will Jagy Mar 8 '16 at 3:09
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    $\begingroup$ It may happen that some $i$ is equal to $p_1$. $\endgroup$ – Unit Mar 8 '16 at 3:14
  • $\begingroup$ i=2p1 (p_i)(p_i - 1) > (p_1 - 1)/p_1 $\endgroup$ – Takahiro Waki Mar 8 '16 at 3:17
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Your error is in the claimed inequality

$$\frac{p_i}{p_i-1}<\frac{p_1-(i-1)}{p_1-i}\;.\tag{1}$$

Take, for example, $x=2\cdot3\cdot5$ and $i=3$: $(1)$ claims that $\frac54<\frac0{-1}=0$, which is absurd. (And taking $i=2$ is in some ways even more of a problem!)

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