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What is Schwarz inequality in $\mathbb R^2$ or $\mathbb R^3$? Give another proof of it in these cases.

Here is my attempt in $\mathbb R^2$. Let $x=(x_1,x_2)$ and $y=(y_1,y_2)$ both in $\mathbb R^2$. The Cauchy-Schwarz inequality is

$$\lvert\langle x,y \rangle \rvert \leq \lVert x\rVert \lVert y\rVert.$$ Then our claim is

$$(x_1y_1+x_2y_2)^2 \leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$

Proof for this inequality: since $(x_1y_2-x_2y_1)^2 \geq 0$, add $(x_1y_1+x_2y_2)^2$ to both sides: \begin{align*} (x_1y_1+x_2y_2)^2 & ≤ (x_1y_2-x_2y_1)^2+(x_1y_1+x_2y_2)^2 \\ & =(x_1y_2)^2+(x_2y_1)^2 -2(x_1y_2)(x_2y_1) \\ & \qquad +(x_1y_1)^2+(x_2y_2)^2+2(x_1y_1)(x_2y_2) \\ & =x_1^2(y_1^2+y_2^2) +x_2^2(y_1^2+y_2^2) \\ & =(x_1^2+x_2^2)(y_1^2+y_2^2), \end{align*} namely $$(x_1y_1+x_2y_2)^2\leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$ This proves the claim.

My question is: is this answer complete to the given question? As there is a choice for $\mathbb R^2$ or $\mathbb R^3$. If any step is missing please identify it.

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  • $\begingroup$ I guess you should do a similar thing in the case of $\mathbb R^3$. The steps look good to me. $\endgroup$ – Gibbs May 28 '20 at 12:10
  • $\begingroup$ Note that I have modified the format of the entire question. Please, use MathJax next times. See here. $\endgroup$ – Gibbs May 28 '20 at 12:22
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Your proof is fine ! But:

you wrote: $ \quad "$ our claim is

$$ (x_1y_1+x_2y_2)^2 \leq (x_1+x_2)(y_1+y_2)."$$

It should read:

$$(x_1y_1+x_2y_2)^2 \leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$

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