$x,y,z>0$, prove:$\frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$ If $x,y,z>0$, prove:$$ \frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$$ I tried to use classical ineqalities such as AM-GM or QM-GM to prove this inequality. I have noticed equality holds when $x=y=z$.
The next part of the question is asking if this inequality can be generalized.

Another inequality with same nature.If $x,y,z>0$, prove or disprove:$$ \frac{x}{y+z+\sqrt[6]\frac{y^6+z^6}{2}}+\frac{y}{z+x+\sqrt[6]\frac{z^6+x^6}{2}}+\frac{z}{x+y+\sqrt[6]\frac{x^6+y^6}{2}}\geq1$$
  These questions are proposed from R. Shahbazov and J. Hajimir to RMM.

 A: Since $$\sqrt[4]{\frac{x^4+y^4}{2}}\leq\sqrt{x^2-xy+y^2}$$ it's just $(x-y)^4\geq0,$ it's enough to prove that
$$\sum_{cyc}\frac{z}{x+y+\sqrt{x^2-xy+y^2}}\geq1$$ or
$$\sum_{cyc}\frac{z(x+y-\sqrt{x^2-xy+y^2})}{3xy}\geq1$$ or
$$\sum_{cyc}(x^2y+x^2z-xyz)\geq\sum_{cyc}z^2\sqrt{x^2-xy+y^2}$$ or
$$\sum_{cyc}\left(\frac{x^2y+x^2z}{2}-xyz\right)\geq\sum_{cyc}z^2\left(\sqrt{x^2-xy+y^2}-\frac{x+y}{2}\right)$$ or
$$\sum_{cyc}z(x-y)^2\geq\sum_{cyc}\frac{3z^2(x-y)^2}{2\sqrt{x^2-xy+y^2}+x+y}$$ or
$$\sum_{cyc}z(x-y)^2\left(1-\frac{3z}{2\sqrt{x^2-xy+y^2}+x+y}\right)\geq0.$$
Now, let $x\geq y\geq z$.
Thus, $$y^2(x-z)^2\geq x^2(y-z)^2,$$ $$2\sqrt{x^2-xz+z^2}\geq2x-z,$$ $$2\sqrt{y^2-yz+z^2}\geq2y-z,$$ $$1-\frac{3y}{2\sqrt{x^2-xz+z^2}+x+z}\geq1-\frac{3y}{2x-z+x+z}\geq0$$ and 
$$y^2\sum_{cyc}z(x-y)^2\left(1-\frac{3z}{2\sqrt{x^2-xy+y^2}+x+y}\right)\geq$$
$$\geq y\cdot y^2(x-z)^2\left(1-\tfrac{3y}{2\sqrt{x^2-xz+z^2}+x+z}\right)+ x\cdot y^2(y-z)^2\left(1-\tfrac{3x}{2\sqrt{y^2-yz+z^2}+y+z}\right)\geq$$
$$\geq y\cdot x^2(y-z)^2\left(1-\tfrac{3y}{3x-z+x+z}\right)+ x\cdot y^2(y-z)^2\left(1-\tfrac{3x}{2y-z+y+z}\right)=$$
$$=xy(y-z)^2\left(x\left(1-\frac{y}{x}\right)+y\left(1-\frac{x}{y}\right)\right)=0$$ and we are done! 
