Saudi Arabia Mathematical Competitions

With the notation $d=\frac{1}{abc}$ the inequality becomes $$\frac{1}{a+b+d+1}+\frac{1}{b+c+d+1}+\frac{1}{c+a+d+1}+\frac{1}{a+b+c+1}\le 1.$$ Let $x=\sqrt[4]{a}$, $y=\sqrt[4]{b}$, $z=\sqrt[4]{c}$, $w=\sqrt[4]{d}$. Then $$xyzw=\sqrt[4]{abcd}=1$$ and it suffices to prove that $$\begin{array}{c}\sum \frac{1}{x^4+y^4+z^4+zyxw}\le 1.\end{array}\text{\hspace{1em}(1)}$$ But $x^4+y^4+z^4\ge x^2{y}^2+y^2{z}^2+z^2{x}^2$ so $$\begin{array}{rl}{x}^4+y^4+z^4& =\frac{x^4+y^4+z^4+x^2{y}^2+y^2{z}^2+z^2{x}^2}{2}\\ & =\frac{x^4+y^2{z}^2}{2}+\frac{y^4+z^2{x}^2}{2}+\frac{z^4+x^2{y}^2}{2}\\ & \ge x^{2}yz+y^{2}zx+z^{2}xy=xyz(x+y+z)\end{array}$$ where we used the AM-GM inequality three times. The same result follows from the rearrangement inequality. Hence $$\begin{array}{c}\frac{1}{x^4+y^4+z^4+xyzw}\le \frac{1}{xyz(x+y+z)+xyzw}\\ =\frac{1}{xyz(x+y+z+w)}=\frac{w}{x+y+z+w}\end{array}$$ and so $$\sum \frac{1}{x^4+y^4+z^4+xyzw}\le \sum \frac{w}{x+y+z+w}=1$$