How find the numbers of $n$ such $\sigma{(n)}\varphi{(n)}>n^2-n$ find all positive integer numbers $n\ge 2$, such that
$$\sigma{(n)}\varphi{(n)}>n^2-n$$
where $\varphi{(n)}$ and $\sigma{(n)}$ be the Euler function of n and the sum of divisors function of $n$,respectively.
I know $\varphi{(n)}\le n$,and I know $$\sigma{(n)}\varphi{(n)}<n^2$$ ,But at last I can't 
 A: Things become - in my opinion - a little nicer to write if we divide the inequality by $n^2$ to obtain
$$\frac{\sigma(n)}{n}\cdot \frac{\varphi(n)}{n} > 1 - \frac{1}{n}.$$
Both, $\sigma$ and $\varphi$ are multiplicative functions, and therefore $n \mapsto \frac{\sigma(n)}{n}\cdot \frac{\varphi(n)}{n}$ is multiplicative too, so it is helpful to know how that behaves for prime powers. For a prime power $p^k$, we have
$$\sigma(p^k) = \sum_{j=0}^k p^j = \frac{p^{k+1}-1}{p-1}, \quad \varphi(p^k) = p^{k-1}(p-1),$$
and therefore
$$\frac{\sigma(p^k)}{p^k}\cdot \frac{\varphi(p^k)}{p^k} = \frac{(p^{k+1}-1)p^{k-1}}{p^{2k}} = \frac{p^{k+1}-1}{p^{k+1}} = 1 - \frac{1}{p^{k+1}}.$$
Thus, for $n \geqslant 2$, we obtain
$$\frac{\sigma(n)\varphi(n)}{n^2} = \prod_{p\mid n}\left(1 - \frac{1}{p^{v_p(n)+1}}\right),\tag{1}$$
where $v_p(n)$ is the multiplicity of $p$ in the prime factorisation of $n$.
All factors on the right hand side of $(1)$ are smaller than $1$, so to have
$$\frac{\sigma(n)\varphi(n)}{n^2} > 1 - \frac1n,$$
we must have
$$1 - \frac{1}{p^{v_p(n)+1}} > 1 - \frac{1}{n}\tag{2}$$
for all primes $p$ dividing $n$. By elementary manipulations, $(2)$ is equivalent to
$$p^{v_p(n)+1} > n,$$
or
$$p > \frac{n}{p^{v_p(n)}}.\tag{3}$$
If $n$ is not a prime power, then $(3)$ cannot hold if $p$ is the smallest prime dividing $n$, since then $\frac{n}{p^{v_p(n)}}$ is the product of powers of primes larger than $p$ and is $> 1$.
For $n$ a prime power, on the other hand, $(3)$ becomes $p > 1$, which is true.
Hence: For an integer $n \geqslant 2$, we have
$$\sigma(n)\varphi(n) > n^2 - n$$
if and only if $n$ is a prime power.
