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.
 A: 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 :
$$(b+c)\sqrt{b^2+c^2-bc} \geq 2 \sqrt{bc} \cdot \sqrt{bc}=2bc$$ 
A: From the Law of Sines, we know $\frac{\sin\alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin\gamma}{c}$, so
$$ \frac{1}{b}+\frac{1}{c} = \sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\frac{1}{a}.$$
It suffices to prove
$$\sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge 2.$$
Since $\alpha = \frac{\pi}{3}$, we have $\beta+\gamma = \frac{2\pi}{3}$, and so $\sin\alpha = \sin\left(\frac{\beta+\gamma}{2}\right)$. As such, it suffices to prove
$$\sin\left(\frac{\beta+\gamma}{2}\right)\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge 2 \iff \frac{1}{2}\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge\frac{1}{\sin\left(\frac{\beta+\gamma}{2}\right)} $$
which follows from Jensen's inequality, since $\frac{1}{\sin}$ is convex on $(0,\pi)$.
A: $$a^2=b^2+c^2-2bc\cos\frac{\pi}{3}\iff a^2+bc=b^2+c^2$$ ►Is it $a^2\ge bc$?
YES because if not, $a^2=bc-\epsilon=b^2+c^2-bc$ with $\epsilon >0$; it follows $-\epsilon=(b-c)^2$, absurde.
Now from $AM\ge GM$ one has $$\frac 1c+\frac 1b\ge 2\sqrt{\frac{1}{bc}}$$ But   $$\sqrt{\frac{1}{bc}}\ge \frac 1a\iff a^2\ge bc$$
Thus $$\frac 1c+\frac 1b\ge\frac2a$$
