Let $z = x + y$. Then $0 \le z \le 1$, whence $z(1-z) \ge 0$. Now
$$\begin{aligned} 5x(1-x) + 5y(1-y) - 8xy &= 5(x+y) - 5(x^2+y^2) - 8xy \\ &= 5z - 5(x^2+2xy+y^2) + 2xy \\ &= 5z - 5z^2 + 2xy \\ &= 5z(1-z) + 2xy \\ &\ge 0, \end{aligned}$$

Solution 2:

From $x+y \le 1$ we obtain $y \le 1-x$ and $x \le 1-y$. As $x \ge 0$ and $y \ge 0$, these imply $xy \le x(1-x)$ and $xy \le y(1-y)$. Multiplying both by 5 and adding them we obtain
$$10xy \le 5x(1-x) + 5y(1-y).$$
As $xy \ge 0$, we have $8xy \le 10xy$ which implies the result.
For equality we need $8xy = 10xy$ and $10xy = 5x(1-x) + 5y(1-y)$. Hence $xy = 0$, and $xy = x(1-x)$ as well as $xy = y(1-y)$. Therefore, we either have $x = 0, y = 1$ or $y = 0, x = 1$ or $x = y = 0$.