If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$. If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$.
Since $\alpha$ is pi/3 that means $ b\le a \le c $.
Then i used AM and HM ,so $\frac 1c+\frac 1b\ge\frac {4}{c+b} $ .
and since $\alpha$ is pi/3 and  $ b\le a \le c $ then $ \frac b2 + \frac c2 \le a $ ,then we replace c+b with 2a and get what we wanted.
This is my work i think it's not a correct proof( or it's ok?),any idea how(else?) to prove this.

From cosine's theorem : 
$$a^2=b^2+c^2-2bc \cos \alpha=b^2+c^2-bc$$
The inequality is equivalent with :
$$a(b+c) \geq 2bc$$ or equivalently :
$$(b+c)\sqrt{b^2+c^2-bc} \geq 2bc$$
Now use the known inequalities $$b+c \geq 2 \sqrt{bc}$$ and $$b^2+c^2 \geq 2bc$$ to get :
$$(b+c)\sqrt{b^2+c^2-bc} \geq 2 \sqrt{bc} \cdot \sqrt{bc}=2bc$$ 
