Show that $\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2$ 
Show that 
  $$\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2$$

I have tried:
$$\vec a \cdot \vec b = {1 \over 4} |\vec a + \vec b|^2 - {1 \over 4} |\vec a - \vec b|^2
\\ \vec a \cdot \vec b = {1\over 4}(|\vec a|^2+2|\vec a||\vec b|+|\vec b|^2) - {1\over 4}(|\vec a|^2 -2|\vec a||\vec b|+|\vec b|^2)
\\ \vec a \cdot \vec b = |\vec a||\vec b|$$
However it seems that the equation only holds true when $\vec a$ and $\vec b$ are collinear.
Is that true? Or did something go wrong?
Thanks

HINT
Such formula is known as the Polarization Identity. Due to the inner product properties, we have:
\begin{align*}
\lVert\textbf{a}+\textbf{b}\rVert^{2} = \langle\textbf{a}+\textbf{b},\textbf{a}+\textbf{b}\rangle = \langle\textbf{a},\textbf{a}\rangle + 2\langle\textbf{a},\textbf{b}\rangle + \langle\textbf{b},\textbf{b}\rangle = \lVert\textbf{a}\rVert^{2} + 2\langle\textbf{a},\textbf{b}\rangle + \lVert\textbf{b}\rVert^{2}
\end{align*}
Analogously, we have
\begin{align*}
\lVert\textbf{a}-\textbf{b}\rVert^{2} = \langle\textbf{a}-\textbf{b},\textbf{a}-\textbf{b}\rangle = \langle\textbf{a},\textbf{a}\rangle - 2\langle\textbf{a},\textbf{b}\rangle + \langle\textbf{b},\textbf{b}\rangle = \lVert\textbf{a}\rVert^{2} - 2\langle\textbf{a},\textbf{b}\rangle + \lVert\textbf{b}\rVert^{2}
\end{align*}
Can you proceed from here?
