Kind of confused what is being asked here. Isn't this just obviously true? How would I start this proof?
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$\begingroup$You can show this as follows;
expanding $||x+y||^2=\langle x+y,x+y \rangle=||x||^2+2\langle x, y \rangle + || y||^2$
if you want to say why this is, then it is just from the properties of inner product. i.e., $$\langle x+y , x+y \rangle= \langle x, x+y \rangle + \langle y, x+y \rangle$$ (linearity)
$$= \langle x+y , x \rangle + \langle x+y , y \rangle$$ (symmetric property)
$$= \langle x , x \rangle + \langle y , x \rangle + \langle x , y \rangle + \langle y ,y \rangle$$
(Linearity again)
and similarly
$|| x-y || ^2= \langle x-y,x-y \rangle= ||x|| ^2 - 2 \langle x, y \rangle + || y||^2$
Now just add these two equations to obtain the final result.
ie, $$||x+y||^2+||x-y||^2=2||x||^2+2||y||^2$$
In terms of a geometric interoperation in $\mathbb{R^2}$ think of two of the sides being the vector x and the other two sides being the vector y. The vectors $x+y$ and $x-y$ correspond to the diagonals.
Photo from "Schaums linear algebra, 4th edition"
$\endgroup$ 1 $\begingroup$When proving the parallelogram law, yes prove it by expanding the norms and cancelling.
For the diagram, think about the corner of the parallelogram a the origin. Define the two neighbouring vertices as $a$ and $b,$ and the fourth as $a+b.$
The left and right sides of the parallelogram have length $|b|$ and the top and bottom sides of the parallelogram have length $|a|.$
If we express the diagonals in terms of $a$ and $b.$ Doing ths, the sum of the square of the lengths of the diagonals is $|a+b|^2+|b-a|^2$ and the sum of the square of the lengths of the sides is $2|a|^2+2|b|^2,$ and the law comes from the fact tat we are showing that these are equal.
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