Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$ 
Let $x,y,z>0$ and $x+y+z=1$. Prove that
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$

My work so far:
I use Titu's Lemma:
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=$$
$$=\frac{x^4}{xy}+\frac{y^4}{yz}+\frac{z^4}{zx}+\frac{x^2}{\frac zx}+\frac{z^2}{\frac yz}+\frac{y^2}{\frac xy}\ge$$
$$\ge\frac{\left(x^2+y^2+z^2+x+y+z\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}=\frac{\left(x^2+y^2+z^2+1\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}$$
I need help here.

Suppose $x \geq y \geq z$. Using the rearrangements inequality we have
$$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq x^3 \cdot \frac{1}{x}+...+z^3 \cdot \frac{1}{z} =x^2+y^2+z^2 $$
since $x^3,y^3,z^3$ and $1/x,1/y,1/z$ have opposite ordering. The same thing applies to 
$$ \frac{x^3}{z}+\frac{z^3}{y}+\frac{y^3}{x} \geq x^2+y^2+z^2$$
Thus the left hand side is greater than $2(x^2+y^2+z^2)$. This is immediately greater than the RHS by noting that $1=(x+y+z)^2$.

You could use "Titu's Lemma" if you like, just like in your try, but split in half:
$$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq \frac{(x^2+y^2+z^2)^2}{xy+yz+zx} \geq \frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2} = x^2+y^2+z^2 $$
