Proof of $\sum_{n=1}^\infty a_nb_n^2c_n^3<\infty$.

Question

Let $a_n \geq 0, b_n \geq 0, c_n \geq 0 \; \forall n \in \mathbb{N}$ and suppose that $\sum_{n=1}^\infty a_n^3b_n^3, \;\sum_{n=1}^\infty b_n^3c_n^3, \; \sum_{n=1}^\infty c_n^6$ are convergent. Prove that the series $\sum_{n=1}^\infty a_nb_n^2c_n^3$ is convergent.

I saw that $(a_n^3b_n^3 \cdot b_n^3c_n^3 \cdot c_n^6)^\frac{1}{3}=a_n b_n^2 c_n^3$ and here I stopped.
I was thinking about using Cauchy-Schwarz inequality, but I don't really know how to apply it (especially having $x_n^3$, not $x_n^2$).

I would be grateful for some hints! 🙂

Answer 1

Apply the Arithmetic Mean - Geometric Mean inequality: $$\frac{a_i^3b_i^3 +b_i^3c_i^3+c_i^6}3\geqslant \sqrt[3]{a_i^3b_i^6c_i^9}=a_ib_i^2c_i^3$$ So your sequence is bounded.