Question 176
For any real numbers $x, y$ prove the inequality
$$
\sqrt{(x + 4)^2 + (y + 2)^2} + \sqrt{(x - 5)^2 + (y + 4)^2} \le \sqrt{(x - 2)^2 + (y - 6)^2} + \sqrt{(x - 5)^2 + (y - 6)^2} + 20.
$$
$$
\sqrt{(x + 4)^2 + (y + 2)^2} + \sqrt{(x - 5)^2 + (y + 4)^2} \le \sqrt{(x - 2)^2 + (y - 6)^2} + \sqrt{(x - 5)^2 + (y - 6)^2} + 20.
$$