Estimating $\prod_{p\mid n}(1+1/p)$ in terms of n

Let $n\geq 1$ be an integer, $P$ the set of rational prime numbers. I am interested in upper bounds for $\prod_{p\in P; \ p\mid n}(1+1/p)$ in terms of n.

I would like to find explicit real numbers $a,b$ such that for any integer $n\geq 1$ it holds $\prod_{p\in P; \ p\mid n}(1+1/p)\leq a\log(n)^b$.

By the answer of GH this is possible for any $b>0$ provided $n$ is sufficiently large.

Now, I would like to find explicit real numbers $a,c$ such that for any integer $n\geq 10$ it holds $\prod_{p\in P; \ p\mid n}(1+1/p)\leq a\log(n)+c$.

What are small possible values for $c$ and $a$?

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If you want a weaker bound that is completely explicit, the following method is pretty simple. Let $\omega(n)$ denote the number of distinct prime factors of $n$. Since the function $1+1/x$ is a decreasing function of $x$, and the smallest prime equals 2, we certainly have $$\prod_{p\mid n} \bigg( 1+\frac1p \bigg) \le \prod_{k=2}^{\omega(n)+1} \bigg( 1+\frac1k \bigg) = \frac{\omega(n)+2}2$$ since the product telescopes. Moreover, $\omega(n) \le (\log n)/(\log 2)$, again since each prime factor of $n$ is at least 2. Therefore $$\prod_{p\mid n} \bigg( 1+\frac1p \bigg) \le \frac{\log n}{2\log2}+1$$ (with equality at $n=1$ and $n=2$).

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This completely answers my question, thank you very much. –  Ping Chen Mar 8 '13 at 20:51
It is easy to verify that for sufficiently large $n$ $$\sum_{p\mid n}\frac{1}{p}<\sum_{p<2\log n}\frac{1}{p}<\log\log\log n+O(1)$$ whence $$\prod_{p\mid n} \left( 1+\frac{1}{p} \right) \ll \log\log n.$$ In other words, you can choose any $b>0$. Also, one can further refine the above bound.
Indeed, for any constant $C>6e^\gamma/\pi^2$ (where $\gamma$ is Euler's constant), we have $$\prod_{p\mid n}\bigg(1+\frac1p\bigg) < C\log\log n$$ for all $n>n_0(C)$. And this is best possible, in that there are infinitely many violations if $C<6e^\gamma/\pi^2$. One proof uses the identity $$\prod_{p\mid n} \bigg(1+\frac1p \bigg) = \frac n{\phi(n)} \prod_{p\mid n} \bigg(1-\frac1{p^2} \bigg)$$ together with known lower bounds for $\phi(n)$. –  Greg Martin Mar 8 '13 at 20:25