$1. (x+y+z)^4+0^4+0^4 \geq x^4+y^4+z^4$ with equality at $y=0$ and $z=0$ So, maximum is at $(x,y,z)=(1+\sqrt 13,0,0)$ $2.$ $ x^4+y^4+z^4 \geq x^4+(\frac{y+z}{2})^4+(\frac{y+z}{2})^4$ So, minimum is at $(x,y,z)=(2,2,2)$

galav wrote: $1. (x+y+z)^4+0^4+0^4 \geq x^4+y^4+z^4$ What if $(x,y,z)=(-1,0,1)$? @below: oops

The Q says positive. Plus- It is same as maximizing and minimizing $x+y+z$

rkm0959 wrote: Let $x,y,z \ge 0$ be real numbers such that $(x+y-1)^2+(y+z-1)^2+(z+x-1)^2=27$. Find the maximum and minimum of $x^4+y^4+z^4$ Write the condition as $$x^2+y^2+z^2+(x+y+z)^2-4(x+y+z)=24.\qquad (1)$$Note that $$x^2+y^2+z^2-4(x+y+z)=(x-2)^2+(y-2)^2+(z-2)^2-12\geq -12,$$then from $(1)$ we obtain $(x+y+z)^2\leq 36,$ i.e., $x+y+z\leq 6$. Hence $$(x+y+z)^2-4(x+y+z)\leq 2(x+y+z) \leq \dfrac{x^2+4}{2}+\dfrac{y^2+4}{2}+\dfrac{z^2+4}{2}=\dfrac{x^2+y^2+z^2+12}{2}.$$Combined with $(1)$, we get $$x^2+y^2+z^2+\dfrac{x^2+y^2+z^2+12}{2}\geq 24,$$$$x^2+y^2+z^2\geq 12.\qquad (2)$$Now, note that $x+y+z\geq \sqrt{x^2+y^2+z^2}$ and $x+y+z+\sqrt{x^2+y^2+z^2}\geq 2\sqrt{x^2+y^2+z^2}>4$, we obtain $$ (x+y+z-\sqrt{x^2+y^2+z^2})(x+y+z+\sqrt{x^2+y^2+z^2}-4)\geq 0,$$$$ (x+y+z)^2-4(x+y+z) \geq x^2+y^2+z^2-4\sqrt{x^2+y^2+z^2}.$$Combined with $(1)$, we get $$ 2(x^2+y^2+z^2)-4\sqrt{x^2+y^2+z^2} \leq 24,$$$$2\left(\sqrt{x^2+y^2+z^2}-1\right)^2\leq 26,$$$$ \sqrt{x^2+y^2+z^2}\leq \sqrt{13}+1$$$$ x^2+y^2+z^2 \leq (\sqrt{13}+1)^2.\qquad (3)$$From $(2)$ and $(3)$, we get the following inequalities $$ (\sqrt{13}+1)^2 \geq x^2+y^2+z^2\geq 12.$$Now, since $$ (x^2+y^2+z^2)^2\geq x^4+y^4+z^4\geq \dfrac{(x^2+y^2+z^2)^2}{3},$$we obtain $$ (\sqrt{13}+1)^4 \geq x^4+y^4+z^4\geq 48.$$For $x=\sqrt{13}+1,\ y=z=0$, we have $x^4+y^4+z^4= (\sqrt{13}+1)^4$. For $x=y=z=2$, we have $x^4+y^4+z^4=48$. Hence, $\max\{x^4+y^4+z^4\}=(\sqrt{13}+1)^4$ and $\min\{x^4+y^4+z^4\}=48$. $\square$

Very nice. $\left(\sum_{cyc}x\right)^2 =36-\sum_{cyc}\left(x-2\right)^2\leq 36 \implies 0\leq\sum_{cyc}x\leq6,$ $\sum_{cyc}x^2 =24-\left(\sum_{cyc}x\right)^2 +4\sum_{cyc}x\geq 24-2\sum_{cyc}x\geq12\implies \sum_{cyc}x^4 \geq \dfrac{\left(\sum_{cyc}x^2\right)^2}{3}\geq48.$ $\sum_{cyc}x^2 =\left(\sqrt{13+\sum_{cyc}xy}+1\right)^2-2\sum_{cyc}xy =14+2\sqrt{13+\sum_{cyc}xy}-\sum_{cyc}xy\leq14+2\sqrt{13}$ $\implies\sum_{cyc}x^4\leq\left(\sum_{cyc}x^2\right)^2\leq\left(\sqrt{13}+1\right)^4.$ Let $x,y,z \ge 0$ be real numbers such that $(x+y-1)^2+(y+z-1)^2+(z+x-1)^2=3$. Find the maximum and minimum of $x^4+y^4+z^4$