Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(x^2 + y^2)^2}$, then substitute $(x^2 + y^2)=t$
$1\geq \lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}\geq \lim_{t\to0}t^{\frac{1}{4}t^2}=\lim_{t\to0}e^{\frac{1}{4}t^2\ln t}=e^0=1$
This is how my professor solved this limit. What I don't understand is this part:
$\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$
How can I prove it? And this would never come to my mind, is there maybe some other way to solve the limit? Grateful in advance.

Here's another approach to this limit. Notice for any $(x,y)\neq (0,0)$ that $$
x^2y^2 \ln(x^2+y^2)=f(x,y)(x^2+y^2)\ln(x^2+y^2)$$ where $f(x,y)=\frac{x^2y^2}{x^2+y^2}$. Clearly $(x^2+y^2)\ln(x^2+y^2)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ while $f(x,y)$ is bounded on the punctured disc $x^2+y^2<1$, $(x,y)\neq (0,0)$. To see this, observe $f(x,0)=0$ for $x\in(-1,0)\cup(0,1)$ and for $x^2+y^2<1,y\neq 0$ we have $$\bigg|\frac{x^2y^2}{x^2+y^2}\Bigg|\leq \Bigg|\frac{x^2y^2}{0+y^2}\Bigg|=x^2\leq x^2+y^2<1$$ This shows $f$ is bounded above by $1$ on the punctured disc, making $$\lim_{(x,y)\rightarrow (0,0)} x^2y^2\ln(x^2+y^2)=0$$ Finally, $$\lim_{(x,y)\rightarrow (0,0)}(x^2+y^2)^{x^2y^2}=\lim_{(x,y)\rightarrow (0,0)}e^{x^2y^2\ln(x^2+y^2)}=e^0=1$$
