furthurmore we have: $ \frac {x^3}{y^3 + 8} + \frac {y^3}{z^3 + 8} + \frac {z^3}{x^3 + 8}\geq\frac {1}{3}\geq\frac {1}{9} + \frac2{27}(xy + xz + yz)$ $ \sum \frac {x^3}{y^3 + 8} \geq \frac {{(x^3 + y^3 + z^3)}^2}{(x^3y^3 + y^3z^3 + z^3x^3) + 8(x^3 + y^3 + z^3)}$$ \geq\frac {{(x^3 + y^3 + z^3)}^2}{\frac {1}{3}{(x^3 + y^3 + z^3)}^2 + 8(x^3 + y^3 + z^3)}$. $ \frac {{(x^3 + y^3 + z^3)}^2}{\frac {1}{3}{(x^3 + y^3 + z^3)}^2 + 8(x^3 + y^3 + z^3)}\geq \frac {1}{3}\longleftrightarrow$ $ (x^3 + y^3 + z^3)\geq3$ you can use couchy and also generalized jensen for convexity of $ \frac {1}{x}$ to prove first inequality

$ \frac{x^3}{(y+2)(y^2-2y+4)}+\frac{y+2}{27}+\frac{y^2-2y+4}{27} \ge x/3$ Summing up cyclically, we have $ \frac{x^3}{y^3+8}+\frac{y^3}{z^3+8}+\frac{z^3}{x^3+8}+\frac{x^2+y^2+z^2-(x+y+z)+6*3}{27} \ge 1$ blah $ \ge 1/3+1/9-\frac{x^2+y^2+z^2}{27}$ Hence it suffices to show that $ 1/3-\frac{x^2+y^2+z^2}{27} \ge (2/27)(xy+xz+yz)$ $ 9-(x^2+y^2+z^2) \ge 2(xy+xz+yz)$ $ 9 \ge (x+y+z)^2=9$ Q.E.D.

Another solution http://www.mathlinks.ro/viewtopic.php?t=223062

My solution is same as you roffler ,but we can multiple the second and third term in main inequality of solution by x.^3 and proving inequality by AM-GM .

Using Hölder Inequality we get : $(\sum x +2)(\sum x^{2}-2x+4)(\sum \frac{x^{3}}{y^{3}+2}) \geq (\sum x)^{3}=27$. Hence we get :$LHS \geq \frac{3}{15-2q} \geq \frac{1}{9}+\frac{2}{27}q$ where $q=\sum xy$. The last inequality is obviously true since it's equivalent to :$(q-3)^{2} \geq 0$ Q.E.D

sumita wrote: furthurmore we have: $ \frac {x^3}{y^3 + 8} + \frac {y^3}{z^3 + 8} + \frac {z^3}{x^3 + 8}\geq\frac {1}{3}$ The following inequalities are also true. Let $ x,y,z\in\mathbb R^{+}$ and $ x+y+z=3$. Prove that\[ \frac {y^3}{x^3 + 8} + \frac {z^3}{y^3 + 8} + \frac {x^3}{z^3 + 8}\geq\frac {1}{3}.\] \[ \frac{y^3+8}{x^3}+\frac{z^3+8}{y^3}+\frac{x^3+8}{z^3} \geq27.\]