Show that if $n$ is composite, then $\phi(n) \leq n-\sqrt{n}$

Question

Please help me showing this:

If $n$ is composite, then $\phi(n) \leq n-\sqrt{n}$.

I failed to proceed from the definition of Euler function $\phi(n)$. First of all if $n$ is composite, then it means that it can be written as a product of prime numbers that are unique. So that is what I know so far and I don't know how to proceed from that point.

Answer 1

Hint(one of my best solutions):

1. If we can write $n=ab$ with $a$ and $b$ are coprime $\varphi(n)=\varphi(a)\varphi(b)$ and $\varphi(a)\leq a-1$ and $\varphi(b)\leq b-1$ and conclude that $$\varphi(n)\leq ab-a-b+1\leq ab-\sqrt{ab}$$
2. If we cannot write $n=ab$ with $a$ and $b$ coprime then $n=p^k$ for som $k$ and $p$ a prime but in this case: $$\varphi(p^k)=p^k-p^{k-1}\leq p^k-p^{\frac{k}{2}} $$ because $k> 1$