How to use Cauchy-Scharw inequality to prove continuity of a function? I'm attempting to understand how to prove the function f such that $$f(x,y)=\frac{x^3y}{x^4+y^2}\;if\;(x,y)\neq (0,0)$$ $$f(x,y)=(0,0)\;if\;(x,y)=(0,0)$$
is continuous in $\mathbb R^2$.
The solution provided says to use the Cauchy-Schwarz inequality to show that $|x^3y|=|x^2yx|\le \frac{(x^2)^2+y^2}{2}|x|$, and thus the function is lesser than or equal to $\frac{|x|}{2}$, and thus the limit goes to 0 as (x,y) goes to to (0,0).  I understand the limit part, but not the use of the inequality.  How do I use Cauchy-Schwarz to obtain this inequality
 A: This inequality is actually one of the standard inequalities and often stated in form of
$$
\vert ab\vert \leq \frac{1}{2} a^2 + \frac{1}{2} b^2,
$$
which directly follows from
$$
0 \leq \frac{1}{2}(\vert a\vert-\vert b\vert)^2 = \frac{1}{2} a^2 - \vert ab \vert + \frac{1}{2} b^2 \\
\Rightarrow \vert ab\vert \leq \frac{1}{2} a^2 + \frac{1}{2} b^2
$$
In your example we have $a = x^2$ and $b = \vert y\vert$.
There is an even more general version to it in form of
$$
 \vert ab\vert \leq \frac{1}{p} a^p + \frac{1}{q} b^q
$$
for $\frac{1}{p} + \frac{1}{q} = 1$. This is called Young's inequality (see Wikipedia).
So I'm kind of surprised that you're actually referenced to the Schwarz inequality here. The funny thing, however, is that I just today happened to read a paper where just this inequality was also called Schwarz inequality. They probably haven't meant Cauchy-Schwarz, though. Maybe there is some historical reason that some people attribute this inequality to Schwarz (and I don't even know if that would be the Schwarz from Cauchy-Schwarz then).
