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While reading I came across a fact that : $$\sqrt{n}/2 ≤ φ(n) ≤ n$$ The upper bound is obvious but how to prove the lower bound on the toitent function ? Thanks

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1 Answer 1

In fact you can improve the lower bound to $\sqrt{n/2}$. Consider

$$\frac{\varphi(n)}n=\prod_k\frac{p_k-1}{p_k}\;,$$

where the $p_k$ are the distinct prime factors of $n$. We also have

$$\frac1{\sqrt n}\le\prod_k\frac1{\sqrt{p_k}}\;.$$

For all primes $p$ we have

$$ \frac{p-1}p\ge\frac1{\sqrt{2p}}\;, $$

with equality only for $p=2$. The lower bound follows.

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