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I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,

Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$

Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$

We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$

As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$

I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof?

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Showing two quantities are less than some other quantity does not show that they are equal e.g. $0 \leq 2$ and $1 \leq 2$ but $0$ is certainly not $1$!

Instead, for this problem, use the following "quadratic formula" for norms (essentially generalizing $(a + b)^2 = a^2 + 2ab + b^2$) If $x, y \in \mathbb{R}^n$ where $n \in \mathbb{Z}_+$, then because the dot product behaves a lot like regular multiplication we have: \begin{align*} ||x + y||^2 &= (x + y) \cdot (x + y) \\ &= (x + y)\cdot x + (x + y)\cdot y \quad \text{ ($\cdot$ distributes over $+$) } \\ &= ||x||^2 + y \cdot x + x \cdot y + ||y||^2 \quad \text{ ($||z||^2$ is just $z \cdot z$) } \\ &= ||x||^2 + 2(x \cdot y) + ||y||^2 \quad \text{ ($\cdot$ commutes) } \end{align*} Using this, can you complete?

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Why don't you just use the definition of the norm $||x||^2=x\cdot x$. Then, starting from the right hand side: $$\begin{align}\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2&=\frac 14(u+v)\cdot(u+v)-\frac 14(u-v)\cdot(u-v)\\&=\frac 14(u\cdot u+2u\cdot v+v\cdot v)-\frac 14(u\cdot u-2u\cdot v+v\cdot v)\\&=\frac14(4u\cdot v)\\&=u\cdot v \end{align}$$

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What's left is to prove, that $\frac{1}{4}(||u-v||^2+||u+v||^2) \geq ||u||\ ||v||$.

We can use triangle inequality one more time:

$\frac{1}{4}(||u-v||^2+||u+v||^2) \geq \frac{1}{4}||u-v-u-v||^2$

and

$\frac{1}{4}(||-u+v||^2+||u+v||^2) \geq \frac{1}{4}||-u+v-u-v||^2$

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No, your proof is inadequate for the reason you stated. If I say $0 \le x \le 1$ and $0 \le y \le 1$, this tells you nothing about whether $x = y$, $x < y$, or $x > y$. Similarly, demonstrating that the RHS of your identity is bounded above by some expression, and then showing the LHS obeys the same bound, tells you nothing about whether the two sides of the identity are in fact equal.

The identity you wish to prove is called the polarization identity. For a complete proof, refer to the following Wikipedia page: Polarization identity (Wikipedia)

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Just simplify the RHS , using dot product properties.

see: http://chortle.ccsu.edu/vEctorlessons/vch07/vch07_8.html.

But if you want to use inequality's, you can show RHS $ \ge $ LHS, and RHS $ \le $ LHS. Then RHS=LHS.

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