Problem (by Baoqian Liu): Let $x, y, z \ge 0$ with $x+y+z > 0$. Prove that \begin{align} &\sum_{\mathrm{cyc}} \sqrt[4] {[2(x+y)^2+2(y+z)^2-(x+z)^2][2(y+z)^2+2(x+z)^2-(x+y)^2]}\\ \ge \ & \frac{9}{4}\sum_{\mathrm{cyc}} \sqrt{\frac{x(x+y)(x+z)(x+y+z)}{(2x+y+z)^2}} + \frac{63}{8}\sqrt{\frac{xyz}{x+y+z}}. \end{align} Clearly, when $x=y=z$, the inequality holds with equality. Also, if $x=y$ and $z=0$, the inequality holds with equality.
My attempt: WLOG, assume that $x+y+z = 3$ (homogeneity) and $x\ge y\ge z$ (symmetry).
If $z \ge (\frac{2}{5})^4 = 0.0256$: Since $2(x+y)^2+2(y+z)^2-(x+z)^2 = 4y(x+y+z)+(z-x)^2 \ge 4y(x+y+z)$ and $(x+y)(x+z) \le \frac{(2x+y+z)^2}{4}$, it suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt[4] {[4y(x+y+z)][4z(x+y+z)]} \ge \frac{9}{8}\sum_{\mathrm{cyc}} \sqrt{x(x+y+z)} + \frac{63}{8}\sqrt{\frac{xyz}{x+y+z}}$$ or $$6(\sqrt[4]{xy}+\sqrt[4]{yz}+\sqrt[4]{zx}) \ge \frac{27}{8}(\sqrt{x}+\sqrt{y}+\sqrt{z}) + \frac{63}{8}\sqrt{xyz}.$$ Let $x=a^4, y=b^4, z=c^4$. It suffices to prove that, for $a\ge c, b\ge c, c\ge \frac{2}{5}$ with $a^4+b^4+c^4=3$, $$48(ab+bc+ca) \ge 27(a^2+b^2+c^2) + 63a^2b^2c^2.$$ This inequality is true. I verified it by Mathematica.
Any comments are welcome.