Is $\arctan(y/x) > xy/(x^{2}+y^{2})$ true for positive $x,y$? I am trying to prove the following:
$$\arctan\frac{y}{x}>\frac{xy}{x^{2}+y^{2}},\quad\forall x,y>0$$
Is the statement true, and if so how do you prove it?
 A: Hint:
Consider the function $h:y\mapsto \arctan\tfrac yx-\tfrac{xy}{x^2+y^2}$. What is $h'(y)$? Is it strictly positive? What can you conclude then?
A: First substitute $t=y/x$, and establish
$$\arctan(t)>\frac t{1+t^2}.$$
We have
$$\arctan(0)=\frac 0{1+0^2},$$ and by differentiation,
$$\frac1{1+t^2}>\frac{1-t^2}{(1+t^2)^2}=\frac1{1+t^2}\frac{1-t^2}{1+t^2}.$$
The last fraction is obviously bounded by $1$, so that the property holds for all $t>0$.
A: Setting $y/x=u\gt0$, the inequality to prove becomes
$$\arctan u\gt{u\over1+u^2}$$
For $u\gt0$, we have
$$\arctan u=\int_0^u{dt\over 1+t^2}\gt\int_0^u{dt\over1+u^2}={u\over 1+u^2}$$
since $1+t^2\lt1+u^2$ for $0\lt t\lt u$.
Alternatively, rewrite the inequality as
$${1\over2}(x^2+y^2)\arctan(y/x)\gt{1\over2}xy$$
and reinterpret the left hand side as ${1\over2}r^2\theta$ where $r=\sqrt{x^2+y^2}$ and $\tan\theta=y/x$.  The left hand side is thus the area of a circular wedge that contains a right triangle of height $y$ and base $x$ (with hypotenuse $\sqrt{x^2+y^2}$), whose area is expressed by the right hand side.
A: Write $y=\sin \theta$, $x=\cos \theta$, then we need to prove: $\theta > \sin \theta \cos \theta$ in $(0,\pi/2)$, multiplying by 2, $x> \sin x$ for $x \in (0, \pi)$ which is true.
