<s>[INST] Inequality for a two variables function I have tried to solve this exercise but I am having difficulties.
I have to find a function greater than $$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}}$$ as the parameter $\alpha$ changes.
That's what I tried:
Since $$sin(xy)\leq1$$
$$\frac{|y|^\alpha \sin(xy)}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{(x^2+y^2)^{3/2}}$$
Then:
$$(x^2+y^2)^{3/2} \geq y^3 $$ so that:
$$\frac{|y|^\alpha}{(x^2+y^2)^{3/2}} \leq \frac{|y|^\alpha}{y^{3}}$$
Now I have problems and I don't know what I did wrong. Following my reasoning I should have:
$$\frac{|y|^\alpha}{y^{3}}=|y|^{\alpha - 3}$$
But I Know it's wrong.
 [/INST] $x^2, y^2 \geq 0$ thus $x^2 + y^2 \geq 0$ and $x^2+y^2\geq y^2$ because $x^2 \geq 0$.
Then put both sides to the power $\frac{3}{2}$
$(x^2+y^2)^\frac{3}{2}\geq (y^2)^\frac{3}{2}=|y|^3$ instead of $y^2$
and we can do this because $y^2 \geq 0$ and $x^2 + y^2 \geq 0$ and $\frac{d}{dz}[z^\frac{3}{2}]=\frac{3}{2}z^\frac{1}{2}\geq 0$ for $z\geq 0$, i.e. $z^\frac{3}{2}$ is an increasing function, so if $z_2\geq z_1$ then $z_2^\frac{3}{2} \geq z_1^\frac{3}{2}$.
Other than that, it doesn't look like the typo really affected your answer.
</s>