TITLE: Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: $P=\sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\dfrac{24}{5\sqrt{5a+5b}}$
$m=\sqrt{\dfrac{a}{1-a}},n=\sqrt{\dfrac{b}{1-b}} \implies a=\dfrac{m^2}{m^2+1},b=\dfrac{n^2}{n^2+1},\\a\ge b\ge c \implies a\ge \dfrac{1}{3} \implies      m^2\ge \dfrac{1}{2} ,a+b=1-c \ge 1-b \implies n\ge \sqrt{\dfrac{1}{2m^2+1}}\implies  mn\ge \sqrt{\dfrac{m^2}{2m^2+1}}=\sqrt{\dfrac{1}{2+\dfrac{1}{m^2}}}\ge \dfrac{1}{2} \implies 4mn \ge 2 $
$\dfrac{1}{a+b}=\dfrac{m^2n^2+m^2+n^2+1}{2m^2n^2+m^2+n^2}\ge \dfrac{(\dfrac{m+n}{2})^2+1}{2(\dfrac{m+n}{2})^2} \iff (n-m)^2(n^2+m^2+4mn-2) \ge 0$
$(\dfrac{m+n}{2}) =\dfrac{1}{t} \implies P \ge \dfrac{2}{t}+\dfrac{24}{5\sqrt{10}}\sqrt{1+t^2}=2f(t) \\ f'(t)=-\dfrac{1}{t^2}+\dfrac{12}{5\sqrt{10}}\sqrt{\dfrac{t^2}{1+t^2}}=0 \implies t^2=\dfrac{5}{3}$
it is easy to verify this is the min point.
$f_{min}=\dfrac{39}{5\sqrt{15}}, P_{min}=\dfrac{78}{5\sqrt{15}}$ when $m=n$
so $m=n=\dfrac{1}{t} \implies a=b=\dfrac{1}{1+t^2}=\dfrac{3}{8}>\dfrac{1}{3}$