How is $\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$ $$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \leq \frac{\sqrt{|xy|}}{\sqrt{2}}$$
Does this apply in general, because here in this example i have $x \to 0, y \to 0$. Some inequality is used I believe to prove this one, but I do not see which.
 A: $$\begin{align*}
(|x|-|y|)^2 &\ge 0\\
x^2 - 2|xy| + y^2 &\ge 0\\
x^2 + y^2 &\ge 2|xy|\\
\sqrt{x^2+y^2} &\ge \sqrt{2|xy|}\\
\frac{1}{\sqrt2} &\ge \frac{\sqrt{|xy|}}{\sqrt{x^2+y^2}}\\
\frac{\sqrt{|xy|}}{\sqrt2} &\ge \frac{|xy|}{\sqrt{x^2+y^2}}\\
\end{align*}$$
A: by $AM-GM$ we get $$\frac{x^2+y^2}{2}\geq |xy|$$ this is equivalent to
$$\frac{1}{\sqrt{x^2+y^2}}\le \frac{1}{\sqrt{2}\sqrt{|xy|}}$$
multiplying by$$|xy|>0$$ we get what we want
A: A proof without using AM-GM inequality. 
Every point $(x, y)$ in $\mathbb{R}^2$ has its polar representation
$$x = r\cos \theta, \quad y = r \sin \theta.$$
Therefore, the inequality to be shown is equivalent to 
$$\left|\frac{r^2\cos\theta\sin\theta}{r}\right| \leq \frac{r\sqrt{|\cos\theta\sin\theta|}}{\sqrt{2}}.$$
Simplifying, it is
$$\sqrt{|2\cos\theta\sin\theta|} \leq 1$$
Or
$$\sqrt{|\sin(2\theta)|} \leq 1$$
which is obviously true.
A: This inequality is equivalent to $ \sqrt{2|xy|} \le \sqrt{x^2 + y^2}$. Squaring this inequality, you get $2|xy| \le x^2 + y^2$ which is true, because $0 \le (|x| - |y|)^2 = x^2 + y^2 - 2|xy| $.
A: Alternatively, let $ABC$ be a right triangle with $\angle ACB=\frac{\pi}{2}$, $BC=|x|$, and $CA=|y|$.  Let $D\in AB$ be the foot of the perpendicular from $C$, and $E\in AB$ the point such that $CE$ internally bisects the angle $ACB$.  Then, $CD=\frac{|xy|}{\sqrt{x^2+y^2}}$ and $CE=\frac{\sqrt{2}|xy|}{|x|+|y|}$.  By AM-GM, $|x|+|y|\geq2\sqrt{|xy|}$, so $CE\leq\sqrt{\frac{|xy|}{2}}$.  As $CD$ is the shortest distance from $C$ to the line $AB$, we have
$$\frac{|xy|}{\sqrt{x^2+y^2}}=CD\leq CE\leq\sqrt{\frac{|xy|}{2}}\,.$$
The equality holds iff $D=E$, or $|x|=|y|$.
A: Try starting from the inequality between arithmetic and geometric means of two (positive, real) numbers:
$$
\frac{x + y}{2} \geq \sqrt{xy}\,.
$$
This in turn follows from the identity $(x-y)^2 \geq 0$:
$$
(x-y)^2 \geq 0 \Leftrightarrow x^2 + y^2 \geq 2 xy \Leftrightarrow x^2 + 2xy + y^2 \geq 4xy \Leftrightarrow \Big(\frac{x+y}{2}\Big)^2 \geq xy.
$$
Can you manipulate this in the a form like the inequality you present?
A: We have $x^2+y^2 \geq 2|xy|$ or $\sqrt{x^2+y^2}\geq \sqrt{2} \sqrt{|xy|}$, or $\frac{1}{\sqrt{x^2+y^2}} \leq \frac{1}{\sqrt{2}\sqrt{|xy|}}$ 
Now if one multiplies both sides with $|xy|$ one gets the inequality.
