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 $x+y\in (0,2]$?
 A: 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.
A: Remarks: @Q. Zhang gave a nice proof. I give an alternative proof here.

Fact 1: $f(v) = v^v - v^2$ is strictly decreasing on $(0, 1)$.
The proof is given at the end.
Fact 2: $u^u \ge \frac{u^2 - u + 2}{3 - u}$ for all $u$ in $(0, 2)$.
The proof is given at the end.
(Note: $\frac{u^2 - u + 2}{3 - u}$ is the Pade $(2,1)$ approximant of $u^u$ at $u = 1$.)
If $x, y \le 1$, we have $x^x \ge x \ge x^2$
and $y^y \ge y \ge y^2$. The desired inequality is true.
In the following, WLOG, assume $x \in (1, 2)$.
By Fact 1, since $0 < y \le 2 - x < 1$, we have $y^y - y^2 \ge (2 - x)^{2 - x} - (2 - x)^2$.
It suffices to prove that, for all $x\in (1, 2)$,
$$x^x + (2 - x)^{2 - x} - x^2 - (2 - x)^2 \ge 0.$$
Using Fact 2, we have
\begin{align*}
 &x^x + (2 - x)^{2 - x} - x^2 - (2 - x)^2\\
 \ge\ &  \frac{x^2 - x + 2}{3 - x} + \frac{(2 - x)^2 - (2 - x) + 2}{3 - (2 - x)} - x^2 - (2 - x)^2\\
 =\ & \frac{2(x - 1)^4}{(3 - x)(1 + x)}\\
  \ge\ & 0.
\end{align*}
We are done.

Proof of Fact 1: We have $f'(v) = v^v(\ln v + 1) - 2v$.
If $v \in (0, \mathrm{e}^{-1}]$, then clearly $f'(v) < 0$. If $v\in (\mathrm{e}^{-1}, 1)$, then
$f'(v) \le \ln(v) + 1 - 2v \le (v - 1) + 1 - 2v = -v < 0$ where we have used $v^v \le 1$ and $\ln v \le v - 1$. We are done.
Proof of Fact 2: Let
$F(u) = u\ln u - \ln \frac{u^2 - u + 2}{3 - u}$.
We have
\begin{align*}
 F'(u) &= \ln u - \frac{(u - 1)(u^2 - 4u + 7)}{(3 - u)(u^2 - u + 2)},\\
 F''(u) &= \frac{(u^4 - 5u^3 + 3u^2 - 19u + 36)(u - 1)^2}{u(3 - u)^2(u^2 - u + 2)^2}.
\end{align*}
Let $h(u) = u^4 - 5u^3 + 3u^2 - 19u + 36$.
We have $h'(u) = u^2(4u - 15) + (6u - 19) < 0$.
Thus, $h(u)$ is strictly decreasing on $(0, 2)$.
Also, $h(1) > 0$ and $h(2) < 0$. Thus,
there exists $u_0\in (1, 2)$ such that $h(u_0) = 0$,
$h(u) > 0$ on $(0, u_0)$, and $h(u) < 0$ on $(u_0, 2)$. Thus, $F''(u)\ge 0$ on $(0, u_0]$
and $F''(u) < 0$ on $(u_0, 2)$.
Thus, $F(u)$ is convex on $(0, u_0]$ and concave on $(u_0, 2)$.
Since $F(1) = 0$ and $F'(1) = 0$, we have
$F(u)\ge 0$ on $(0, u_0]$. Also, since $F(u_0)\ge 0$ and $F(2) = 0$,
we have $F(u)\ge 0$ on $(u_0, 2)$.
Thus, $F(u)\ge 0$ on $(0, 2)$. We are done.
