<s>[INST] 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 $x+y\in (0,2]$?
 [/INST] 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.
</s>