# Proving an inequality (with Cauchy-Schwarz?)

Let $(a_j)$ and $(b_j)$ be non-negative numbers, and for $k\geq 0$ define $c_k=\sum _{j=0}^ka_jb_{k-j}$. I'm trying to show that : $$\sum _{k=0}^{\infty}\frac{c_k^2}{k+1}\leq (\sum _{j\geq 0}a_j^2)(\sum _{j\geq 0}b_j^2).$$

By Cauchy-Schwarz we have $c_k^2\leq (\sum _{j= 0}^ka_j^2)(\sum _{j=0}^kb_j^2)\leq(\sum _{j\geq 0}a_j^2)(\sum _{j\geq 0}b_j^2)$, but the harmonic series diverges, so we can't prove the inequality this way.

By homogeneity we can assume that $\sum _{j\geq 0}a_j^2=1=\sum _{j\geq 0}b_j^2$. It remains to show that $\sum _{k=0}^{\infty}\frac{c_k^2}{k+1}\leq 1$ but I don't see how to proceed for the moment. Any idea will be welcomed.

Using Cauchy product formula we have:

$(\sum_{n=0}^{\infty}a_n^2)(\sum_{n=0}^{\infty}b_n^2)=(\sum_{n=0}^{\infty}d_n)$, where:

$d_n=\sum_{k=0}^{n} a_k^2b_{n-k}^2$

Notice that if we show that:

$\frac{c_n^2}{n+1} \leq d_n$

The inequality will be proven, but it's Cauchy-Schwarz inequality for:

$x_k=1$

$y_k=a_k b_{n-k}$, where $k=0,1 \cdots, n$