It is known about numbers $$a\_1, a\_2, ... a\_n$$ that $$a\_1 + a\_2 +...+a\_n \le 1/2.$$ Prove that $$ (1-a\_1)(1-a\_2)...(1-a\_n) \ge 1/2$$
I have tried using $a^2 \geq 0$, it led to nothing.
How can I make my inequality look like in the possible duplicate?