45th Mongolian Mathematical Olympiad

It's easy to check, when $a=1$, $b=2$, $c=3$ equality holds, hence by the Cauchy's inequality we have $$\begin{array}{rl}& \frac{1}{4}\left(2\sqrt{\frac{3b}{a+b}\cdot \frac{4c}{a+b+c}}+4\sqrt[4]{\frac{3a}{a+b}{\left(\frac{2(a+b)}{a+b+c}\right)}^2}\cdot 1\right)\le \\ & \le \frac{1}{4}\left(\frac{3b}{a+b}+\frac{4c}{a+b+c}+\frac{3a}{a+b}+\frac{2(a+b)}{a+b+c}+\frac{2(a+b)}{a+b+c}+1\right)=\\ & =\frac{1}{4}(3+4+1)=2.\end{array}$$ Here equality holds for if and only if $\frac{3b}{a+b}=\frac{4c}{a+b+c}$ same as $\frac{3a}{a+b}=\frac{2(a+b)}{a+b+c}=1$. Thus equality holds for $a>0$, $b=2a$, $c=3a$.