show that if $ n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n) \leq n/3$ Show that if $n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n)$ is less than or equal to $n/3$ 
I tried a few cases
If $n=6$, $n-1=5$, and $n+1=7$ then $~\phi(6)=2=n/3$
If $n=12$, $n-1=11$, and $n+1=13$ then  $~\phi(12)=4=n/3$
$\phi(n)=n(1-1/p_1)(1-1/p_2)...(1-1/p_k)$
 A: You need to show that $$\phi(n)\leq \frac{n}{3}$$
If $n>4$, with $n-1$ and $n+1$ both being primes,
Now since $n-1$ is a prime $p =(n-1)> 3$ 
And therefore odd, it follows that, $$(n-1)+1=n, \text{ is even}$$
Likewise we know that at least one of $3$ successive integers, must be divisible by $3$.
Now looking at the integers $\{ n-1,n,n+1 \}$ we see that $n-1$ and $n+1$ are primes
So there only divisors are $1$ and themselves
Thus if $3$ divides $n-1$ or $n+1$, we must have either $3=n-1$ or that $3=n+1$
But we know that $n-1>3$ and $n+1>5$, so we see neither of these can happen 
And thus $3$ must divide $n$
So we have that $3\mid n$ and $2\mid n$
Now we need to show that $$\phi(n)\leq \frac{n}{3}$$
Or equivalently $$\frac{\phi(n)}{n} \leq \frac{1}{3}$$
Or $$\prod_{p\mid n}(1-\frac{1}{p})\leq \frac{1}{3}$$
But since $2\mid n$ and $3\mid n$ we can re-write this as,
$$\frac{1}{3}\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})=(1-\frac{1}{3})(1-\frac{1}{2})\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq \frac{1}{3}$$
$$\iff \frac{1}{3}\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq \frac{1}{3}$$
$$\iff \prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq 1$$
Which is clearly true,
Thus if $n>4$, with $n-1$ and $n+1$ both being primes
$$\phi(n)\leq \frac{n}{3}$$
As required.
