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Let $V$ be the real inner product space. Prove that for every $x, y \in V$, it holds $||x+y||^2 + ||x-y||^2 = 2(||x||^2 + ||y||^2)$

Maybe I don't have a lot of knowledge in inner product space . I have read the definition and work at some problems in my textbook, but I can't solve this one unfortunately. Could you help me?

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  • $\begingroup$ This has been discussed here previously. It is known as the parallelogram identity or the parallelogram law. Knowing what it is called should help you find resources for a proof (it just requires the definition of the norm-squared as an inner product and properties of the inner product). $\endgroup$ – hardmath Mar 14 '18 at 5:05
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    $\begingroup$ Start with $\|v\|^2=v\cdot v$. $\endgroup$ – Gerry Myerson Mar 14 '18 at 7:42
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$$\|x+y\|^2+\|x-y\|^2=$$ $$=(x+y)\cdot (x+y)+(x-y)\cdot (x-y)=$$ $$= (x+y)\cdot x +(x+y)\cdot y\;+\;(x-y)\cdot x -(x-y)\cdot y=$$ $$= x\cdot x +y\cdot x +x\cdot y +y\cdot y\;+\;x\cdot x-y\cdot x-x\cdot y +y\cdot y=$$ $$=2(x\cdot x +y\cdot y)=2(\|x\|^2+\|y\|^2).$$

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  • $\begingroup$ Not leaving much for OP to do. $\endgroup$ – Gerry Myerson Mar 14 '18 at 12:11

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