Inequality from Israel TST Let $a, b, c, d$ be nonnegative numbers such that $a+b+c+d=18.$
Prove that:
$$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$
These are my attempts to solve the Problem:
There is an equality case in $(a, b, c, d) = (8, 1, 8, 1)$ and similar cyclic permutations.
To preserve the equality case we'll use Cauchy that way:
$$\sum_{cyc}\sqrt{\frac{a}{b+6}}=\sum_{cyc}\sqrt{\frac{a}{a+6}\cdot\frac{a+6}{b+6}}\leq\sqrt{\sum_{cyc}\frac{a}{a+6}\sum_{cyc}\frac{a+6}{b+6}}$$
Which preserves the equality case. I don't know how to proceed now.
Any ideas?
 A: Let $\{a,b,c,d\}=\{x,y,z,t\}$ such that $x\geq y\geq z\geq t$.
Thus, since $(\sqrt{x},\sqrt{y},\sqrt{z},\sqrt{t})$ and $\left(\frac{1}{\sqrt{t+6}},\frac{1}{\sqrt{z+6}},\frac{1}{\sqrt{y+6}},\frac{1}{\sqrt{x+6}}\right)$ are same ordered, 
by Rearrangement, AM-GM and by Jensen for $f(x)=\sqrt{x+12}$ and for $g(x)=-\frac{1}{\sqrt{x+12}}$ 
we obtain:
$$\sum_{cyc}\sqrt{\frac{a}{b+6}}\leq\sqrt{\frac{x}{t+6}}+\sqrt{\frac{y}{z+6}}+\sqrt{\frac{z}{y+6}}+\sqrt{\frac{t}{x+6}}=$$
$$=\sum_{cyc,x\rightarrow y,t\rightarrow z}\left(\sqrt{\frac{x}{t+6}}+\sqrt{\frac{t}{x+6}}\right)=\sum_{cyc}\sqrt{\frac{x}{t+6}+\frac{t}{x+6}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\sum_{cyc}\sqrt{\frac{x+t}{6}+\frac{x}{t+6}-\frac{x}{6}+\frac{t}{x+6}-\frac{t}{6}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\sum_{cyc}\sqrt{\frac{x+t}{6}-\frac{xt(x+t+12)}{6(x+6)(t+6)}+2\sqrt{\frac{xt}{(x+6)(t+6)}}}=$$
$$=\frac{1}{\sqrt6}\sum_{cyc}\sqrt{x+t-\frac{xt(x+t+12)}{(x+6)(t+6)}+2\sqrt{\frac{36xt}{(x+6)(t+6)}}}\leq$$
$$\leq\frac{1}{\sqrt6}\sum_{cyc}\sqrt{x+t-\frac{xt(x+t+12)}{(x+6)(t+6)}+\frac{xt(x+t+12)}{(x+6)(t+6)}+\frac{36}{x+t+12}}=$$
$$=\frac{1}{\sqrt6}\sum_{cyc}\sqrt{\frac{(x+t)^2+12(x+t)+36}{x+t+12}}=\sum_{cyc}\frac{x+t+6}{\sqrt{6(x+t+12}}=$$
$$=\sum_{cyc}\frac{x+t+12-6}{\sqrt{6(x+t+12}}=\sum_{cyc}\left(\frac{1}{\sqrt6}\cdot\sqrt{x+t+12}-\sqrt6\cdot\frac{1}{\sqrt{x+y+12}}\right)\leq$$
$$\leq\frac{2}{\sqrt6}\sqrt{\frac{x+t+y+z}{2}+12}-\frac{2\sqrt6}{\sqrt{\frac{x+t+y+z}{2}+12}}=\frac{2\sqrt{21}}{\sqrt6}-\frac{2\sqrt6}{\sqrt{21}}=5\sqrt{\frac{2}{7}}.$$
