Lower bound for sum of Hecke eigenvalues

Question

Let $\lambda$ be weakly multiplicative, $\lambda(n)\geq0$, $p$ prime and $S(x)=\sum_{n\leq x}\lambda(n)\log(\frac{x}{n})$ for real $x$.

How can I show $S(x)\gg \left(\sum_{p\leq \sqrt{x/3}}\lambda(p)\right)^2-\left(\sum_{p\leq \sqrt{x/3}}1\right)$?

Here is the background: The question is coming from IKS, section 3. $\lambda(n)$ are the eigenvalues of a newform $f$ of level $N$. All above sums are chosen such that $(n,N)=1$ or $p\not\mid N$, thus giving multiplicativity.

In Xu, section 3.1 something similar happens. Here the hints $\lambda(p)^2=\lambda(p^2)+1$ (which leads to $\lambda(p)\geq1$ and $|\lambda(n)|\leq\sigma_0(n)$ (divisor function) are given.

Answer 1

I found a way which should do the trick. I think the division by 3 is just a technical detail and 2 would be already sufficient to pull $\log(\frac{x}{x/2})$ out of the sum. You can take $S(x)=\sum_{n\leq x}\lambda(n)\log(x/n)\geq \sum_{pq\leq x/2}\lambda(pq)$. Furthermore we use the formula $\lambda(p)^2=\lambda(p^2)+1$, which is (1.2) in IKS., to estimate the case $p=q$. That's where the subtracted sum of 1's comes from.