A question related to Selberg's sieve theory I am currently reading upon Selberg's sieve and following is an argument that I came across:


Multiplying both sides of $\lambda_{\delta}=\delta \sum_{\substack{d<z
 \\ \delta|d}}\frac{\mu(d/\delta)\mu(d)}{\phi(d)V(z)}$, we get $\begin  {align*} V(z)\lambda_{\delta} &=\delta \sum_{\substack{d<z \\
 \delta|d}}\frac{\mu(d/\delta)\mu(d)}{\phi(d)} 
 \\&=\delta\sum_{t<\frac{z}{\delta}}\frac{\mu(t)\mu(\delta 
 t)}{\phi(\delta t)}\\&= \delta \sum_{\substack{t<\frac{z}{\delta} \\ 
 (t,\delta)=1}}\frac{\mu^{2}(t)\mu(\delta)}{\phi(\delta)\phi(t)}\\&= 
 \mu(\delta)\prod_{p|\delta}\left(1+\frac{1}{p-1}\right)\sum_{\substack{t<\frac{z}{\delta}
 \\ (t,\delta)=1}}\frac{\mu^{2}(t)}{\phi(t)}. \end{align*}$



Taking the absolute value of both sides, we see that
$$|V(z)|\cdot|\lambda_{\delta}|\leq |V(z)|.$$

Here $V(z)= \sum_{\delta<z} \frac{\mu^2 (\delta)}{\phi(\delta)} $, where $\phi$ is the Euler's totient function and $\mu$ is Mobius function.
But I don't see how $ \delta \sum_{\substack{t<\frac{z}{\delta} \\ 
 (t,\delta)=1}}\frac{\mu^{2}(t)\mu(\delta)}{\phi(\delta)\phi(t)}= 
 \mu(\delta)\prod_{p|\delta}\left(1+\frac{1}{p-1}\right)\sum_{\substack{t<\frac{z}{\delta}
 \\ (t,\delta)=1}}\frac{\mu^{2}(t)}{\phi(t)}.$ Also, how does it follow that $|V(z)|\cdot|\lambda_{\delta}|\leq |V(z)|$?
Any help will be extremely useful. Thanks.
 A: Note that
$$
\sum_{\substack{t<z/\delta\\(t,\delta)=1}}{\mu^2(t)\mu(\delta)\over\varphi(\delta)\varphi(t)}={\mu(\delta)\over\varphi(\delta)}\sum_{\substack{t<z/\delta\\(t,\delta)=1}}{\mu^2(t)\over\varphi(t)},
$$
and by the Euler product formula for $\varphi(\delta)$, we have
$$
{1\over\varphi(\delta)}=\delta^{-1}\prod_{p|\delta}\left(1-\frac1p\right)^{-1}=\delta^{-1}\prod_{p|\delta}\left(1+{1\over p-1}\right).
$$
Combining these two identities answers the first question.
To answer the second question, one should start off with $V(z)$:
$$
V(z)=\sum_{d<z}{\mu^2(d)\over\varphi(d)}.
$$
Specifically, we sum over $d<z$ according to the value of $(d,\delta)$. That is,
$$
V(z)=\sum_{k|\delta}\sum_{\substack{d<z\\(d,\delta)=k}}{\mu^2(d)\over\varphi(d)}.
$$
To continue, set $d=tk$, so that we can replace the condition $d<z\wedge(d,\delta)=k$ with $t<z/d\wedge(t,\delta)=1$:
\begin{aligned}
V(z)
&=\sum_{k|\delta}{\mu^2(k)\over\varphi(k)}\sum_{\substack{t<z/d\\(t,\delta)=1}}{\mu^2(t)\over\varphi(t)} \\
&\ge\left(\sum_{k|\delta}{\mu^2(k)\over\varphi(k)}\right)\left(\sum_{\substack{\color{red}{t<z/\delta}\\(t,\delta)=1}}{\mu^2(t)\over\varphi(t)}\right) \\
&=\prod_{p|\delta}\left(1+{1\over p-1}\right)\sum_{\substack{t<z/\delta\\(t,\delta)=1}}{\mu^2(t)\over\varphi(t)}
\end{aligned}
