Show that if $n$ is composite, then $\phi(n) \leq n-\sqrt{n}$ 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.
 A: Hint(one of my best solutions): 


*

*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 $$\varphi(n)\leq ab-a-b+1\leq ab-\sqrt{ab}$$

*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 $k> 1$

A: Similar approach @Elaqqad used:


Corollary:
$$(\forall\;\text{prime}\;p)(\forall n\in\Bbb N) p\mid n\implies p\le\sqrt{n}$$



Proposition:
Let $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, then:
$$\begin{aligned}\varphi(n)&=\prod_{i=1}^k\varphi\left(p_i^{\alpha_i}\right)\\&=\prod_{i=1}^kp_i^{\alpha_i}\left(1-\frac1{p_i}\right)\\&=n\left(1-\frac1{p_1}\right)\left(1-\frac1{p_2}\right)\cdots\left(1-\frac1{p_k}\right)<n,\quad\alpha_i\in\Bbb N\;\forall\; i\in\{1,\ldots,k\}\end{aligned}$$

One argument is:
$$1-\frac1{p_i}<1\;\forall i\in\{1,\ldots,k\}$$
One can also use:
$$n=\sum_{d\mid n}\varphi(d)$$

Let $p$ be the least prime factor of composite $n\in\Bbb N$ $($so we could get the maximum $\frac1{p_i}$$)$.
$$\varphi(n)\le n\left(1-\frac1p\right)\leq n\left(1-\frac1{\sqrt{n}}\right)= n-\sqrt{n}$$
$$\text{Q.E.D.}$$
