# On Cauchy-Schwarz inequality

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.

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

you wrote: $$\quad "$$ our claim is
$$(x_1y_1+x_2y_2)^2 \leq (x_1+x_2)(y_1+y_2)."$$
$$(x_1y_1+x_2y_2)^2 \leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$