Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$ As said in the title, I have to find the minimum of the following:
$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$  
In my very last attempt, I tried to work it out using AM-GM:
Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\frac {a}{\sqrt{a(b+c)}}\ge \frac {\sqrt{2}a}{a+b+c}$$
$$\Rightarrow\sum_{cyc} {\sqrt{\frac a{2(b+c)}}}\ge \frac {\sqrt{2}(a+b+c)}{a+b+c}=\sqrt{2}$$
The problem is, to get the minimum value, $$a=b+c$$
$$b=c+a$$
$$c+a+b$$
or $a=b=c=0$. This is wrong since $a,b,c>0$
And after some calculus I found the minimum value is $\frac 32$ when $a=b=c$.
So what mistake I have made?

By AM-GM $\sum\limits_{cyc}\sqrt{\frac{a}{2(b+c)}}=\sum\limits_{cyc}\frac{\sqrt2a}{2\sqrt{a(b+c)}}\geq\sum\limits_{cyc}\frac{\sqrt2a}{a+b+c}=\sqrt2$, 
which is infimum because we can use $c\rightarrow0^+$ and $b=c$.
