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Let the expression be $P$ .

$P=\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}$

$=\frac{a^{2}}{a\sqrt{50-a}}+\frac{b^{2}}{b\sqrt{50-b}}+\frac{c^{2}}{c\sqrt{50-c}}$

$\geq \frac{(a+b+c)^{2}}{a\sqrt{50-a}+b\sqrt{50-b}+c\sqrt{50-c}}$ by Titu's Lemma,

$= \frac{50^{2}}{a\sqrt{50-a}+b\sqrt{50-b}+c\sqrt{50-c}}$

Define $f(x)=x\sqrt{50-x}$

Note that $f''(x)=-(50-x)^{-\frac{1}{2}}-\frac{x}{4}(50-x)^{-\frac{3}{2}} <0$ for $0<x<50$ . Therefore, $f(x)$ is a concave function.

By Jensen's inequality,

$f(a)+f(b)+f(c) \leq 3f(\frac{a+b+c}{3})$

$\implies \frac{1}{f(a)+f(b)+f(c)} \geq \frac{1}{3f(\frac{a+b+c}{3})}=\frac{1}{3f(\frac{50}{3})}$ .

Therefore, continuing our work from above,

$P \geq \frac{50^{2}}{a\sqrt{50-a}+b\sqrt{50-b}+c\sqrt{50-c}}$

$\geq \frac{50^{2}}{3f(\frac{50}{3})}$ by Jensen's inequality.

$= \frac{50^{2}}{3.\frac{50}{3}.\sqrt{\frac{2.50}{3}}}$

$= 5\sqrt{3} \approx \boxed{8.66}$

Equality holds when $a=b=c=\frac{50}{3}$

Call the expression A, now we choose $S=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ca)$ Now applying Holder's Inequality we get $A^2S\geq (a+b+c)^3$ Easily prove this inequality $2(ab+bc+ca)\leq\frac{2}{3}\big(a+b+c\big)^2$ by Cauchy-Schwarz. So finally $A^2\geq\frac{3(a+b+c)}{2}=75$ $\Leftrightarrow A\geq 5\sqrt{3}\approx 8.66$ The equality holds when $a=b=c=\frac{50}{3}$ . Really nice solution, @ZK LIn