Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$ Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$
Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$

I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c=\dfrac{1}{4}$
And this is my try

Applying AM-GM inequality, we get: $\dfrac{b+c}{a}+\dfrac{5}{3}\ge2\sqrt{\dfrac{5}{3}}.\sqrt{\dfrac{b+c}{a}}$
This implies $\sqrt{\dfrac{a}{b+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3a}{3+2a}$
Similarly, $\sqrt{\dfrac{b}{a+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3b}{3+2b}$
We need to prove that: $2\sqrt{\dfrac{5}{3}}\left(\dfrac{3a}{3+2a}+\dfrac{3b}{3+2b}\right)+\dfrac{24}{5\sqrt{5a+5b}}\ge\dfrac{78}{5\sqrt{15}}$

But I have no idea how to continue. Who can help me or have any other idea?

You can use calculus to find this.
$f(a,b) = \sqrt{\frac{a}{1-a}} + \sqrt{\frac{b}{1-b}} + \frac{24}{5\sqrt{5(a + b)}}$
Now, let us first find the critical point, $(x,y)$, of this function, where $f_a(x,y) = 0$ and $f_b(x,y) = 0$.
Taking partial derivatives and setting them to 0, you will see that the critical point is at $a = b$.
So we now have a 1D function, $g(a) = f(a,a) = 2\sqrt{\frac{a}{1 - a}}+ \frac{24}{5\sqrt{10 a}}$. 
Then differentiate $g$ with respect to $a$, set it equal to 0, and you get that $a = 3/8$, so then $b = 3/8$ and $c = 1/4$.
You can make this more concrete by constructing the Hessian matrix of $f(a,b)$ and verifying that this is indeed the minimum (check that the determinant is >0 and that $f_{aa} > 0$ at this critical point). 
