\frac{n}{φ ( n )} = \prod_{p ∣ n} (1 - \frac{1}{p})^{- 1} .

Let

y = \frac{l o g n}{l o g l o g n} .

The factors with

p \leq y

contribute at most

\prod_{p \leq y} (1 - \frac{1}{p})^{- 1} = C lo g y + O (1) = (C + o (1)) lo g lo g n

p > y

, the number of distinct prime divisors of

n

is at most

lo g n / lo g y

, and hence

lo g p ∣ n p > y \prod (1 - \frac{1}{p})^{- 1} ≪ p ∣ n p > y \sum \frac{1}{p} \leq \frac{lo g n}{y lo g y} = O (\frac{1}{lo g lo g n}) .

Thus the large-prime product is

1 + o (1)

, and

\frac{n}{φ ( n )} \leq (C + o (1)) lo g lo g n .

Taking reciprocals gives, uniformly as

n \to \infty

φ (n) \geq (C^{- 1} + o (1)) \frac{n}{lo g lo g n} .

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