Prove that $\phi (p-1) \leq (p-1)/2$, where $p\geq 3$ is a prime Prove that $\phi (p-1) \leq (p-1)/2$, where $p\geq 3$ is a prime
My idea:
We know that $\phi (p^k) = p^{k-1} (p-1)$. Then to use the $\phi$ is multiplicative. now I have $\phi (p-1)$ in my hand, but then how to complete? could anyone help me, please?
Edit: OP didn't check for $p=2$, the only prime where above $\phi (p-1) \leq (p-1)/2$ holds to fail.
 A: Hint:
Two even numbers can't be coprime.
A: Note that for every $p\geq 3$, $(p-1)$ is an even number. Therefore, by fundamental Theorem of Arithmetic, it will have a unique factorization which contains $2^i$, $i\geq 1$. 
Now, Note that $$\phi(2^i)=2^{i-1}\text{ or }\phi(2^i)=\frac{2^i}{2}$$. 
Therefore, using multipicative nature of $\phi$, we have $$\phi(p-1)=\phi(2^i)\phi(\frac{p-1}{2^i}) \leq \frac{p-1}{2}$$
Case $1$ : $(p-1)$ is an exact power of $2$ i.e. $\frac{p-1}{2^i}=1$
$$\phi(p-1)=\phi(2^i)\phi(1) = \frac{p-1}{2}$$
Case $2$ : $\frac{p-1}{2^i}=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{n}^{\alpha_{n}}$, where $p_{j}\geq 3$ $ \forall j=1,2,...,n$ and $\alpha_{j}\geq 0$ $ \forall j=1,2,...,n$ and alteast one of $\alpha_{j}\geq 1$.
$$\phi(\frac{p-1}{2^i})=\phi(p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{n}^{\alpha_{n}})=\phi(p_{1}^{\alpha_{1}})\phi(p_{2}^{\alpha_{2}})...\phi(p_{n}^{\alpha_{n}}) < \frac{p-1}{2^i}$$
Now, Can you complete Case $2$ and combine it with Case $1$ to get the desired answer?
Note: I have used the convention, $\phi(1)=1$.
A: Hint: If $p$ is odd, then $p-1$ is even.  If $p$ is even, then $p=2$ and $\phi(1)$ is defined to be $1$ by convention and so exceeds $1/2$.
