Prove that if $n>10$ then $\sum_{d\mid n}\phi(\phi(d))<\frac{3}5n$ Prove that if $n>10$ then $$\sum_{d\mid n}\phi(\phi(d))<\frac{3}5n,$$
where $\phi(n)$ is Euler's totient function.
 A: We start with the identity: 
$$n=\sum_{d|n}\phi(d).$$
In order to prove it, just note that right hand side is a multiplicative function and therefore it is enough to check equality for prime power only.
Now the key point is to note that if $d|n$ then $\phi(d)|\phi(n)$ and therefore 
the left hand side of our inequality is the sum $\phi(m)$ where $m$ runs over some divisors of $\phi(n).$ In other words,
$$\sum_{d\mid n}\phi(\phi(d))=\sum_{m|\phi(n)}\phi(m)-S=\phi(n)-S,$$
where $S$ is the sum of those divisors of $\phi(n)$ that are not of the from $\phi(d),$ $d|n.$
Now, if $n=p_1^{\alpha_1}\cdotp_2^{\alpha_2}...\cdot p_k^{\alpha_k}$ then
$\phi(n)=p_1^{\alpha_1}\cdotp_2^{\alpha_2}...\cdot p_k^{\alpha_k}(p_1-1)....(p_k-1)$ and those divisors that come from $\phi(d)$ are all of the form 
$m=p_1^{\beta_1}\cdotp_2^{\beta_2}...\cdot p_k^{\beta_k}\prod_{i}(p_i-1).$ So if $n\ne 2^m$ then divisor $D=\frac{p_1^{\alpha_1-1}\cdotp_2^{\alpha_2-1}...\cdot p_k^{\alpha_k-1}(p_1-1)....(p_k-1)}{2}$ is in $S$ and we can estimate:
$$\phi(n)-S\le \phi(n)/2\le \frac{n}{2}\le \frac{3}{5}n.$$
You are left to check $n=2^m$ which can be easily done directly.
