Boundedness of a function of two variables

Question

Let us consider the function from $\mathbb R^2$ to $\mathbb R$ defined by $\phi(\alpha,\beta) = \frac{\alpha \beta}{\sqrt{\alpha ^4 + \beta ^4}}$ and $0$ otherwise.

If we look at the graph using Geogebra, then it is clearly bounded. However, proving this is the case feels like it requires a large number of cases, and it feels as though this is the incorrect approach.

It is trivial to show that when $\alpha = \beta$, then the function is bounded. Although from this point, it feels like a difficult process to bound the function in a simple way without introducing multiple cases for when $\alpha$ and $\beta$ are positive / negative / different and also when they are different combinations of $| \alpha | < 1$ and $| \beta | < 1$.

Is there a simpler solution to the problem, or would bounding this require all of the cases that I listed above?

Answer 1

We use the fundamental inequality $ab\leq\frac{a^2+b^2}2$ for $a,b\geq0$.

We have \begin{align*} |\phi(\alpha,\beta)|&=\frac{|\alpha| |\beta|}{\sqrt{\alpha ^4 + \beta ^4}}=\frac{\sqrt{|\alpha|^2 |\beta|^2}}{\sqrt{\alpha ^4 + \beta ^4}}=\frac1{\sqrt{\alpha ^4 + \beta ^4}}\cdot\sqrt{|\alpha|^2 |\beta|^2}\\ &\leq\frac1{\sqrt{\alpha ^4 + \beta ^4}}\cdot\sqrt{\frac{|\alpha|^4+|\beta|^4}2}=\sqrt{\frac12},\ \ \ (\alpha,\beta)\neq(0,0). \end{align*}

Answer 2

We may also rewrite $$ \phi(\alpha,\beta) = \frac{\beta/\alpha}{\sqrt{1+(\beta/\alpha)^4}}. $$ This is a function of one variable $t:=\beta/\alpha$, which should be amenable to standard elementary methods.

In fact, let $f(t)=\frac{t}{\sqrt{1+t^4}}$. Then $$ |f(t)| = \frac{1}{\sqrt{t^{-2}+t^2}}. $$ It is enough to show that $t^{-2}+t^2$ is bounded below by a positive constant, which reduces further to analyzing $u^{-1}+u$ for $u>0$. It's elementary to show that $u^{-1}+u$ has a global minimum (for $u>0$) at $u=1$ with value $2$. Backtracking all the way back, we may conclude that $$ |\phi(\alpha,\beta)| = |f(\beta/\alpha)|\le \frac{1}{\sqrt{2}}, $$ with equality exactly when $\alpha=\pm\beta$.

(Of course, $\alpha=0$ or $\beta=0$ are special trivial cases not covered by the above).