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.
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}