If $n$ is composite it has a prime divisor less than or equal to $\sqrt{n}$, and there are at least $\frac{n}{\sqrt{n}}$ numbers less than $n$ divisible by this divisor and thus not relatively prime to n. So $\phi(n) \le n - \sqrt{n}$ with equality iff n is the square of a prime.

Just a minor correction. There are at least $\lfloor n/\sqrt{n}\rfloor$ numbers divisible by that divisor, so we are not quite done. However if there is another prime divisor, then $\phi(n)\le n-\lfloor\sqrt{n}\rfloor-1<n-\sqrt{n}$. In case there is no other prime divisor, then $n=p^k$ ($k\ge2$), $\phi(n)=p^k-p^{k-1}\le p^k-p^{k/2}=n-\sqrt{n}$.

To prove $$\varphi{(n)} \leq n - \sqrt{n} $$ $$n = \displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}, \hspace{0.2cm} \ni P_i > P_j, i < j$$$$\varphi(n) \leq n - \sqrt{n} $$$$\varphi(\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}) \leq (\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}) - \sqrt{\displaystyle \prod_{1 \leq i \leq r}{P_i^{\alpha_i}}} $$$$\displaystyle \prod_{1 \leq i \leq r}\left[{P_i^{\frac{\alpha_i}{2}}{\cdot \left(\frac{P_i - 1}{P_i}\right)}}\right] \leq \displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}} - 1$$$$\displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}}\left[1 - \displaystyle \prod_{1 \leq i \leq r}{\left(\frac{P_i - 1}{P_i}\right)}\right] \geq 1$$As, $$\left(\frac{P_i - 1}{P_1}\right) \cdots \left(\frac{P_r - 1}{P_r}\right) \leq \left(\frac{P_r - 1}{P_r}\right)$$$$\displaystyle\prod_{1 \leq i \leq r}{P_i^{\frac{\alpha_i}{2}}}\left[1 - \displaystyle \prod_{1 \leq i \leq r}{\left(\frac{P_i - 1}{P_i}\right)}\right] \geq P_r^{\frac{\alpha_1 + \cdots + \alpha_r}{2}}\left[1 - \left(\frac{P_r - 1}{P_r}\right)\right] = P_r^{\frac{\alpha_1 + \cdots + \alpha_r}{2} - 1} \geq 1$$ $\textbf{EQUALITY}$ Holds when $n = P^{2}$