Proving Two Inequalities I am trying to prove the following two inequalities using AM-GM
First, I must show that the following inequality must hold :
$$
\frac{x_{1}x_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{x_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{x_{2}^{2}}{x_{2}^{2}+y_{2}^{2}}
$$
and
$$\frac{y_{1}y_{2}}{\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}}\leq\frac{y_{1}^{2}}{x_{1}^{2}+y_{1}^{2}}+\frac{y_{2}^{2}}{x_{2}^{2}+y_{2}^{2}}
$$
My Attempt : Let $x_{1},x_{2},y_{1},y_{2}\in\mathbb{R}$,let $x=\sqrt{x_{1}^{2}+x_{2}^{2}}$ and $y=\sqrt{y_{1}^{2}+y_{2}^{2}}$ it appears that I could pove that $\sqrt{\sum_{i=1}^{2}x_{i}^{2}y_{i}^{2}}\leq xy$ as follows :
$$
xy=\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)\left(\sqrt{y_{1}^{2}+y_{2}^{2}}\right)=\sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+x_{2}^{2}y_{1}^{2}}\geq \sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}}
$$
I noticed that it implies that $\frac{\sqrt{x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}}}{\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)\left(\sqrt{y_{1}^{2}+y_{2}^{2}}\right)}\leq 1$
I am unable to continue I hope someone gives me a hint.
 A: We can even prove Cauchy-Schwartz using AM-GM for $\mathbb{R^n}$. Note that by AM-GM:  $$2 = 1+1 = \sum_{i=1}^{n} \frac{a_i^2}{a_1^2+...+a_n^2} + \sum_{i=1}^{n} \frac{b_i^2}{b_1^2+...+b_n^2} = \sum_{i=1}^{n} \left(\frac{a_i^2}{a_1^2+...+a_n^2} + \frac{b_i^2}{b_1^2+...+b_n^2}\right) \geq \sum_{i=1}^{n} \frac{2a_ib_i}{\sqrt{(a_1^2+...+a_n^2)(b_1^2+...+b_n^2)}}$$. Dividing both sides by $2$ and squaring both sides gives us the desired inequality.   For your inequality, if we let $x = \frac{x_1^2}{x_1^2+y_1^2}, y = \frac{x_2^2}{x_2^2+y_2^2}$, Then we are trying to prove $x+y \geq \sqrt{xy}$, Which follows immediatly from AM-GM as $x+y \geq 2\sqrt{xy}$ (and we see that the original inequality will be true even if we divide the RHS by $2$).
A: So assuming that $x$ is a vector with coordination $(x_1, x_2)$ and $y$ is a vector with coordination $(y_1, y_2)$, then the length of $x$ is $|x| = \sqrt{x_1^2 + x_2^2}$, and the length of $y$ is $|y| = \sqrt{y_1^2 + y_2^2}$. Take the dot product of 2 vectors $x$ and $y$ you will have $x\cdot y = x_1y_1 + x_2y_2$
Noticing that the definition of dot product is $x \cdot y = |x||y|\cos\alpha$ where $\alpha$ is the angle between vector $x$ and $y$, we will have $$\frac{x\cdot y}{|x||y|} = \cos \alpha \leq 1$$ $$\Rightarrow \frac{x_1y_1 + x_2y_2}{\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}} \leq 1$$ and we are done
