Let $\delta=\limsup\frac {\psi (x)}x $ i want to prove that $$\sum_{n\le x}\psi (\frac xn)\le (\delta+\epsilon)x\log x+x\psi (N) $$
Given $\epsilon>0$ there is $N$ such that $x>N$ implies that $\psi(x)\le (\delta+\epsilon)x $
$$\sum_{n\le x}\psi (\frac xn)=\sum_{n\le x/N}\psi (\frac xn)+\sum_{x/N<n\le x}\psi (\frac xn)\le (\delta+\epsilon)x\log x+x\psi (N)-(\delta+\epsilon)x (\log N-C+O (\frac 1x) )$$
But i can't justify that the last line is less than $(\delta+\epsilon)x\log x+x\psi (N) $. Since $ \log N-C+O (\frac 1x) )$ can be negative. Could you help me please?
$$\sum_{n \le x} \psi(x/n) = \sum_{n \le x/N} \psi(x/n) + \sum_{x/N < n \le x} \psi(x/n) $$
$$ \le \sum_{n \le x/N} C_N \frac{x}{n} + x \psi(N)\le C_N x (1+\log (x/N))+ x \psi(N) \\ \le (C_N+\frac{1-\log N}{\log x}) x \log x+x\psi(N) \\ \le (C_N+\frac{1}{\log x}) x \log x+x\psi(N) $$ where $C_N = \sup_{x \ge N} \frac{\psi(x)}{x}$.
Moreover as $N \to \infty $ with $x = N^N$ then $C_N + \frac{1-\log N}{\log x} \to \lim \sup_{x \to \infty} \frac{\psi(x)}{x}$.