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?

Stronger inequality:
With $a, b, c> 0$:  $$\sqrt{\frac{a}{b+ c}}+ \sqrt{\frac{b}{c+ a}}+ \sqrt{\frac{c}{a+ b}}\geq  \sqrt{2+ 2.\frac{abc}{\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )}}$$
Proof:
Thus, we have:
$$\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )+ abc= \left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$
From the assumed inequality, we obtain the equivalent inequality:
$$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}\geq  2\sqrt{\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )}$$
Other way, we have:
$$\sqrt{a\left ( a+ b \right )\left ( a+ c \right )}+ \sqrt{b\left ( b+ c \right )\left ( b+ a \right )}+ \sqrt{c\left ( c+ a \right )\left ( c+ b \right )}= \sqrt{a^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{b^{2}\left ( a+ b+ c \right )+ abc}+ \sqrt{c^{2}\left ( a+ b+ c \right )+ abc}\geq  \sqrt{\left ( a\sqrt{a+ b+ c}+ b\sqrt{a+ b+ c}+ c\sqrt{a+ b+ c} \right )^{2}+ 9abc}= \sqrt{\left ( a+ b+ c \right )^{3}+ 9abc}\geq  4\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$
Done!
