An inequality with a load of variables: $ (1-a_1)(1-a_2)...(1-a_n) \ge 1/2$ 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? 
 A: By induction for $n=1$ it is obivious.
You can consider $b_n = a_n+a_{n+1}$
where one has 
$$a_1, a_2, ... a_n$$ that $$a_1 + a_2 +...+\underbrace{a_n +a_{n+1}}_{b_n}\le 1/2 .$$But $$(1-a_n)(1-a_{n+1}) = (1-a_n-a_{n+1} + a_n a_{n+1} )\geq (1-a_n-a_{n+1})= (1-b_n).$$ 
Then $$(1-a_1)(1-a_2)...(1-a_n)(1-a_{n+1})\ge (1-a_1)(1-a_2)...(1-b_n) \ge 1/2$$
A: Hint: you can consider the inequality for arithmetic and geometric means:
$$\frac{1}{n}\sum_{k=1}^n b_i \geq \sqrt[n]{\prod_{k=1}^n b_i}$$
for any set of non-negative real numbers $\{b_1,\cdots,b_n\}$

Partial attempt:
$$b_i = 1-a_i$$
$$\frac{1}{n}\sum_{k=1}^n (1-a_i) \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$
$$1-\frac{1}{n}\sum a_i \geq \sqrt[n]{\prod_{k=1}^n (1-a_i)}$$
$$lhs > 1-\frac{1}{2n}$$
if product $\leq 1/2$ then rhs not sure to be larger than $\sqrt[n]{\frac{1}{2}}$
Now what to remains is to compare which can be done by calculus comparing a linear function with a power function, treating n as continous variable.

The proof that blue does not overtake red can be left as an exercise.
