A Cauchy–Schwarz-like inequality involving dot products Let $\mathbf{a}_1,\ldots,\mathbf{a}_N$ and $\mathbf{b}_1,\ldots,\mathbf{b}_N$ be vectors in $\mathbb{R}^n$. Prove that: 
\begin{equation}
\bigg(\sum_{i=1}^{N} \mathbf{a}_i^\top\mathbf{a}_i\bigg) \bigg(\sum_{i=1}^{N} \mathbf{b}_i^\top\mathbf{b}_i\bigg)
\geq
N  \sum_{i=1}^N \bigg( (\mathbf{a}_i^\top\mathbf{a}_i) (\mathbf{b}_i^\top\mathbf{b}_i) - (\mathbf{a}_i^\top\mathbf{b}_i)^2 \bigg).
\end{equation}

The inequality doesn't seem to be true. Consider the following example in $\mathbb{R}^2$
\begin{align}
\mathbf{a}_1 = 2\mathbf{e}_1,\ \  \mathbf{a}_2 = \mathbf{e}_2 \ \ \text{ and }  \ \ \mathbf{b}_1 = 2\mathbf{e}_2,\ \  \mathbf{b}_2 = \mathbf{e}_1. 
\end{align}
Observe
\begin{align}
(\mathbf{a}_1^T\mathbf{a}_1+\mathbf{a}_2^T\mathbf{a}_2)(\mathbf{b}_1^T\mathbf{b}_1+\mathbf{b}_2^T\mathbf{b}_2) = (4+1)(4+1)=25
\end{align}
and
\begin{align}
2(\mathbf{a}_1^T\mathbf{a}_1\mathbf{b}_1^T\mathbf{b}_1+\mathbf{a}_2^T\mathbf{a}_2\mathbf{b}_2^T\mathbf{b}_2)=2(16+1) = 34
\end{align}
which means the above inequality doesn't hold. 
