Let $\pi(x)$ the prime-counting function and $\varphi(n)$ denotes the Euler's totient function.
I would like to know if next arithmetic function, that I've defined searching a comparison of such arithmetic function, tends to zero as $N\to\infty$
$$f(N)=\frac{1}{N}\sum_{k=1}^N\frac{\pi(\varphi(k)+N)}{\varphi(\pi(k)+N)}\tag{1}$$
When $N=10^D$, with $D=0,1,2,3$ one gets that $f(N)$ is calculated as, respectively, $1,\approx0.917,\approx0.511$ and $\approx0.370$. Thus our positive function is decreasing in this segment.
Question. Prove or refute $$f(N)\to 0$$ as $N\to\infty$. Many thanks.
I presume that the solution needs the prime number theorem and inequalities for the Euler's totient function.
Since $$C_{2}\frac{n}{\log\left(n\right)}\leq\pi\left(n\right)\leq C_{1}\frac{n}{\log\left(n\right)}$$ where $C_{1},C_{2}>0$ are suitable constants and $$\frac{n}{e^{\gamma}\log\left(\log\left(n\right)\right)+\frac{3}{\log\left(\log\left(n\right)\right)}}<\phi\left(n\right)<n$$ where $\gamma$ is the Euler Mascheroni constant, we have $$\pi\left(\phi\left(k\right)+N\right)\ll\frac{\phi\left(k\right)+N}{\log\left(\phi\left(k\right)+N\right)}\ll\frac{2N}{\log\left(N\right)}$$ and $$\frac{1}{\phi\left(\pi\left(k\right)+N\right)}\ll\frac{\log\left(\log\left(\pi\left(k\right)+N\right)\right)}{\pi\left(k\right)+N}\ll\frac{\log\left(\log\left(2N\right)\right)}{N}$$ so $$\sum_{k\leq N}\frac{\pi\left(\phi\left(k\right)+N\right)}{\phi\left(\pi\left(k\right)+N\right)}\ll\frac{\log\left(\log\left(2N\right)\right)}{\log\left(N\right)}\sum_{k\leq N}1\ll\frac{\log\left(\log\left(2N\right)\right)}{\log\left(N\right)}N$$ then the claim.