Let $N\gg 1$ be a large parameter, which I ultimately want to let tend to infinity. I am reading an old paper of Bourgain, where he claims the lower bound (Equation 2.50, pg. 118)
$$\sum_{q=1}^{N^{1/2}-1}\sum_{{1\leq a < q}\atop{(a,q)=1}}\frac{N^{3}}{q^{2}}\geq c(\log N)N^{3}, \tag{1}$$
where $(a,q)$ denotes the GCD of $a$ and $q$ and $c>0$ is some absolute constant, the value of which I don't care and may change from line to line.
Using the following lower bound for Euler's totient function $\varphi(n)$
$$\varphi(n)>\frac{n}{e^{\gamma}\log\log n + \frac{3}{\log\log n}} \tag{2}$$
and the PNT, I only know how to show that
\begin{align*} \sum_{q=1}^{N^{1/2}-1}\sum_{{1\leq a < q}\atop{(a,q)=1}}\frac{N^{3}}{q^{2}}&\geq c\sum_{{1\leq q<N^{1/2}-1}\atop{q=\mathrm{prime}}}\frac{N^{3}}{q^{2}}\cdot q\\ &+c\sum_{{1\ll q<N^{1/2}-1}\atop{q\neq\mathrm{prime}}}\frac{N^{3}}{q^{2}}\cdot\frac{q}{e^{\gamma}\log\log q+\frac{3}{\log\log q}}\\ &\geq cN^{3}\log\log N \tag{3} \end{align*}
If I knew the asymptotics of the series $\sum_{q=1}^{N^{1/2}-1}\frac{\varphi(q)}{q^{2}}$ or $\sum_{q=1}^{N^{1/2}-1}\frac{1}{\sigma(q)}$, where $\sigma$ is the divisor sum function, then my problem would be solved. However, I have been unable to find such information.
John M. has very kindly provided me with the necessary asymptotics. To see how to get that formula, first recall that $\varphi(n)=\sum_{d\mid n}\mu(d)\frac{n}{d}$, where $\mu$ is the Mobius function, and $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}=\frac{6}{\pi^{2}}$. Then
$$\sum_{n\leq x}\frac{\varphi(n)}{n^{2}}=\sum_{n\leq x}\frac{1}{n^{2}}\sum_{d\mid n}\mu(d)\frac{n}{d}=\sum_{{q,d}\atop{qd\leq x}}\frac{\mu(d)}{d^{2}q}=\sum_{d\leq x}\frac{\mu(d)}{d^{2}}\sum_{q\leq\frac{x}{d}}\frac{1}{q} \tag{4}$$
Using the formula $$\sum_{q\leq k}\frac{1}{k}=\log(k)+\gamma+\epsilon_{k},$$ where $\epsilon_{k}\sim\frac{1}{k}$, and $|\log(y)-\log([y])|=O(1/y)$, we see that
\begin{align*} \mathrm{(4)}&=\sum_{d\leq x}\frac{\mu(d)}{d^{2}}[\log(x)-\log(d)+\gamma+O(\frac{1}{x/d})]\\ &=(\log(x)+\gamma)(\frac{6}{\pi^{2}}-\sum_{d>x}\frac{1}{d^{2}})-\sum_{d=1}^{\infty}\frac{\mu(d)\log(d)}{d^{2}}+\sum_{d>x}\frac{\mu(d)\log(d)}{d^{2}}+O(\frac{\log x}{x})\\ &=\frac{6}{\pi^{2}}(\log(x)+\gamma)-A+O(\frac{\log x}{x}) \end{align*}
$$\sum_{q \leq x} \frac{\phi(q)}{q^2} = \frac{6}{\pi^2}(\gamma + \log x) - A + O\left(\frac{\log x}{x}\right)$$ where $A$ is the constant $$A = \sum_{n=1}^\infty \frac{\mu(n) \log{n}}{n^2}$$ and $\gamma$ is Euler's constant.