Here is the question:
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If $x>0, y>0$ and $x+y=1$, prove that $x^{x}y^{y}+x^{y}y^{x} \le 1$.
Proof: According to the conditions, $ x>0, y>0 $ and $ x+y=1 $, we have
$$1 = x+y = x^{x+y} + y^{x+y} = x^x x^y + y^x y^y$$
Therefore,
$$
\begin{align*}
1 - (x^x y^y + x^y y^x) &= x^x x^y + y^x y^y - x^x y^y - x^y y^x \\
&= x^x (x^y - y^y) - y^x (x^y - y^y) \\
&= (x^y - y^y) (x^x - y^x)
\end{align*}
$$
Since, $x>0, y>0$, so $(x^y - y^y)$ and $(x^y - y^y)$ must have the same sign. That is $(x^y - y^y)(x^y - y^y) \ge 0$.
Therefore,
$$1 - (x^x y^y + x^y y^x) \ge 0$$
So,
$$x^x y^y + x^y y^x \le 1$$
$\blacksquare$
