Prove $x^x+y^y\ge x^2+y^2$ for $x,y>0$ and $x+y\le 2$. We may prove the inequality for $x,y\in (0,1]$.
Note that, for $0<x\le 1$, it holds that
\begin{align*}
 x^x&=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\cdots\\
&\ge1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\\
&\ge x^2.
\end{align*}
Similarily, for $y \in (0,1]$, it holds that $$y^y\ge y^2.$$
Thus$$x^x+y^y\ge x^2+y^2.$$
But how to prove under the condition $x+y\in (0,2]$?

Here is a sketch. As you have shown, it suffices to show that
$$x^x+y^y\ge x^2+y^2$$
if one of $x,y$ is bigger than 1. Assume $x\in (1,2]$, then $y\le 2-x\in (0,1).$ One can check that the function $x^x-x^2$ is decreasing on $(0,1)$ and thus
$$y^y-y^2\ge (2-x)^{2-x}-(2-x)^2.$$
Thus it suffices to show that
$$x^x-x^2+(2-x)^{2-x}-(2-x)^2\ge 0$$
for $x\in (1,2].$
This should not be very hard. Actually, the last inequality holds for all $x\in [0,2]$ (since the left side is symmetric about the line $x=1$) and the left side function in increasing on $(1,2)$.
Edit: about the proof of the last inequality. Consider the function
$$f(x)=x^x-x^2+(2-x)^{2-x}-(2-x)^2, x\in [0,2].$$
We want to show that $f'(x)\le 0$ on $[0,1]$ (and $f'(x)\ge 0$ on $[1,2]$). We have
$$f'(x)=x^x(\ln(x)+1)-(2-x)^{2-x}(\ln(2-x)+1)-4x+4,$$
$$f''(x)=x^x((\ln(x)+1)^2+1/x)+(2-x)^{2-x}((\ln(2-x)+1)^2+1/(2-x))-4.$$
Denote $g(x)=x^x((\ln(x)+1)^2+1/x)$. We will check that $g(x)$ is concave up. Thus by Jensen's inequality, we have $g(x)+g(2-x)\ge 2g((x+2-x)/2)=2g(1)=4$, which shows $f''(x)\ge 0$ for $x\in (0,2)$ and thus $f'(x)$ is increasing on $(0,2)$. Note that $f'(1)=0$, we get the result.
Edit 2: The proof of the function $h(x)=x^x-x^2$ is decreasing on $(0,1)$. We have $h'(x)=x^x(\ln(x)+1)-2x=x(x^{x-1}(\ln(x)+1)-2)$. Let $u(x)=x^{x-1}(\ln(x)+1)$. We can check that $u'(x)=x^{x-1}((\ln(x)+1)^2-\ln(x)/x)>0$ on $(0,1)$ since $\ln(x)<0$ on $(0,1)$. Thus $u(x)$ is increasing on $(0,1)$ and $u(x)\le u(1)=1$ for $x\in (0,1)$. Thus $h'(x)=x(u(x)-2)<0$ for $x\in (0,1)$, which shows $h(x)$ is decreasing on $(0,1)$.
Edit 3: A sketch about the proof $g(x)=x^x((1+\ln(x))^2+1/x)$ is concave up on $(0,2)$. We need to show that $g''(x)>0$. Denote $A(x)=x^x$ and $B(x)=((1+\ln(x))^2+1/x)$, so that $g(x)=A(x)B(x)$. We use the formula
$$g''(x)=(AB)''=A''B+2A'B'+AB''.$$
Note that $A''=AB$ and thus $A''B=AB^2>0$.  Denote $C=1+\ln(x)$. We have $A'=AC$. Thus $2A'B'+AB''=A(2B'C+B'')$. Thus it suffices to show that $2B'C+B''>0$ on $(0,2)$. One can check that
$$2B'C+B''=x^{-3}(4x^2(\ln(x))^2+4x(2x-1)\ln(x)+4x^2-2x+2).$$
It is not hard to show the last expression is positive on $(0,2)$. The details are omitted. Note that $g''(x)>0$ is equivalent to the 4th-derivative of $x^x$ is positive on $(0,2)$. I suspect that there should be a simpler proof.
