Cauchy-Shwarz inequality in vector analysis Vectors $x$ and $y$ are related as follows $$\mathbf{x}+\mathbf{y(x \cdot y)}=\mathbf{a}.$$
Show $$\mathbf{(x \cdot y)}^2=\mathbf{\frac{|a|^2-|x|^2}{2+|y|^2}}$$
I think we need to proceed using Cauchy-Shwarz inequality.
$\mathbf{y(x \cdot y)}=\mathbf{a}-\mathbf{x}$
$\mathbf{y(y \cdot x)(y \cdot x)}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$
$\mathbf{y(y \cdot x)^2}=(\mathbf{a}-\mathbf{x)(x \cdot y)}$
Then, I am lost.

We have $a = x + \def\<#1>{\left<#1\right>}\<x,y>y$, hence
\begin{align*}
  |a|^2 &= \<a,a>\\
        &= \<x + {\<x,y>y, x+ \<x,y>y}>\\
        &= \<x,x> + 2\<x,y>^2 + \<x,y>^2\<y,y>\\
        &= |x|^2 + (2 + |y|^2)\<x,y>^2\\
\iff |a|^2 - |x|^2 &=(2 + |y|^2)\<x,y>^2\\
\iff \<x,y>^2  &= \frac{ |a|^2 - |x|^2 }{2 + |y|^2} 
\end{align*}
