The formula for \(\phi(n)\) (derived here) is:
\[ \phi(n) = n \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right) \]
Let \(p\) be the smallest prime divisor of \(n\). Since \(( 1 - \frac{1}{p_i} ) < 1\):
\[ \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right) \le \left( 1 - \frac{1}{p} \right) \]
This means:
\[\begin{align} \phi(n) &\le n \left( 1 - \frac{1}{p} \right) \\ &\le n - \frac{n}{p} \end{align}\]
Since \(p\) divides \(n\), then there exists integer \(k\) such that \(k=\frac{n}{p}\):
\[ \phi(n) \le n - k\]
Since \(n=\sqrt{n}\sqrt{n}\) and \(n=pk\):
\[\begin{align} \text{either } & \quad p \le \sqrt{n}, \ k \ge \sqrt{n} \\ \text{or } & \quad p \ge \sqrt{n}, \ k \le \sqrt{n} \end{align}\]
Since \(p\) is the smallest prime divisor of \(n\) and \(k\) divides \(n\), then the prime factors of \(k\) may include \(p\) and/or primes higher than \(p\), but not primes lower than \(p\). This means \(p \le k\). This must also mean \(p \le \sqrt{n}\) and \(k \ge \sqrt{n}\). Therefore:
\[\begin{gather} -k \le - \sqrt{n} \\ n-k \le n-\sqrt{n} \\ \phi(n) \le n - k \implies \phi(n) \le n - \sqrt{n} \end{gather}\]