Space with inner product identity

Question

Given a space with inner product, show the next identity

\begin{equation} \|z-x\|^{2}+\|z+y\|^{2}=\frac{1}{2}\|x-y\|^{2}+2\left\|z-\frac{1}{2}(x+y)\right\|^{2} \end{equation}

I started with the parallelogram identity but it doesn't work, help please.

Answer 1

The identity

$$\| \mathbf{\zeta} \|^{2} = \mathbf{\zeta} \cdot \mathbf{\zeta} \;\; \forall \mathbf{\zeta}$$

is the definition of the vector’s magnitude (length) $\| \mathbf{\zeta} \|$.

The dot product (inner product) is commutative and distributive over vector addition.

For the first term of your case, $\| \mathbf{z} - \mathbf{x} \|^{2} = ( \mathbf{z} - \mathbf{x} ) \cdot ( \mathbf{z} - \mathbf{x} ) = \mathbf{z} \cdot \mathbf{z} - 2 \, \mathbf{z} \cdot \mathbf{x} + \mathbf{x} \cdot \mathbf{x}$.

Then the same way you expand the remaining magnitudes, separately for the expression on the left and for the expression on the right.

If you will do it carefully and without mistakes, you will get an identity.