Suppose $n$ is odd natural number. Define $r(n)=\sum_{n_1+n_2+n_3=n} \Lambda(n_1)\Lambda(n_2)\Lambda(n_3)$ and $r'(n)=\sum_{p_1+p_2+p_3=n}(\log p_1)(\log p_2)(\log p_3)$ where $p_1,p_2,p_3$ are restricted to primes.
Prove that $r(n) \sim C(n)n^2$ as $n $ tends to infinity if and only if $r'(n) \sim C(n)n^2$ as $ n$ tends to infinity, where $C(n)=\frac12 \prod_{p\mid n}(1-1/(p-1)^2)\prod_{p|n}(1+1/(p-1)^3)$.
My attempt : let's assume that $r(n) \sim C(n)n^2$ as $n $ tends to infinity which means $\lim_{n \to \infty} r(n)/C(n)n^2=1$. We have to show $\lim_{n \to \infty} r'(n)/C(n)n^2=1$. So we have to show that their behavior is asymptotically equivalent. I am stucked now.
Hint: let $\lambda(m) = \log m$ if $m$ is prime and $\lambda(m)=0$ otherwise. Then \begin{align*} r(n) - r'(n) &= \sum_{n_1+n_2+n_3=n} \bigl( \Lambda(n_1)\Lambda(n_2)\Lambda(n_3) - \lambda(n_1)\lambda(n_2)\lambda(n_3) \bigr) \\ &= \sum_{n_1+n_2+n_3=n} \bigl( \Lambda(n_1) - \lambda(n_1) \bigr) \Lambda(n_2)\Lambda(n_3) \\ &\qquad{}+ \sum_{n_1+n_2+n_3=n} \lambda(n_1) \bigl( \Lambda(n_2) - \lambda(n_2) \bigr) \Lambda(n_3) \\ &\qquad{}+ \sum_{n_1+n_2+n_3=n} \lambda(n_1) \lambda(n_2) \bigl( \Lambda(n_3) - \lambda(n_3) \bigr). \end{align*} This should be amenable to a fairly trivial upper bound simply because $\Lambda(m)-\lambda(m)$ is supported on a very small set of integers.