Taking the <natural logarithm> of the finite <Euler product> and using the <Taylor series>
$$
-\log(1-u)=u+\sum_{k\geq2}\frac{u^k}{k}
$$
gives
$$
\log\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=\sum_{p\leq x}\frac1p
+\sum_{p\leq x}\sum_{k\geq2}\frac1{kp^k}.
$$
The double series converges absolutely, and the <Mertens second theorem> therefore makes the right side
$$
\log\log x+\log C+O\left(\frac1{\log x}\right)
$$
for
$$
C=\exp\left(c+\sum_p\sum_{k\geq2}\frac1{kp^k}\right)>0.
$$
Exponentiating proves
$$
\boxed{\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=C\log x+O(1).}
$$
This is the <Mertens third theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / d / Solution

The prime-factor formula for the <Euler totient function> is
$$
\frac n{\varphi(n)}
=\prod_{p\mid n}\left(1-\frac1p\right)^{-1}.
$$
Let
$$
y=\frac{\log n}{\sqrt{\log\log n}}.
$$
The factors with $p\leq y$ contribute at most
$$
\prod_{p\leq y}\left(1-\frac1p\right)^{-1}
=C\log y+O(1)
=(C+o(1))\log\log n
$$
by the <Mertens third theorem>. For $p>y$, the number of distinct prime divisors of $n$ is at most $\log n/\log y$, and hence
$$
\begin{aligned}
\log\prod_{\substack{p\mid n\\p>y}}\left(1-\frac1p\right)^{-1}
&\ll\sum_{\substack{p\mid n\\p>y}}\frac1p\\
&\leq\frac{\log n}{y\log y}
=O\left(\frac1{\sqrt{\log\log n}}\right).
\end{aligned}
$$
Thus the large-prime product is $1+o(1)$, and
$$
\frac n{\varphi(n)}\leq(C+o(1))\log\log n.
$$
Taking reciprocals gives, uniformly as $n\to\infty$,
$$
\boxed{\varphi(n)\geq\left(C^{-1}+o(1)\right)
\frac n{\log\log n}.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / e / Solution

If $C$ is the constant in the <Mertens third theorem>, then
$$
\varphi(n)\geq\left(C^{-1}+o(1)\right)\frac n{\log\log n}.
$$
Split the prime divisors of $n$ at $y=\log n/\sqrt{\log\log n}$. The small primes contribute at most $(C+o(1))\log\log n$ to $n/\varphi(n)$, while the large primes contribute a factor $1+o(1)$.

Totient lower bound from the Mertens product

{c}

There is a constant $C>0$ such that
$$
\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=C\log x+O(1).
$$
Taking logarithms reduces the result to the <Mertens second theorem>, because the terms of order $p^{-2}$ and smaller form an absolutely convergent series.

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