If $x+y+z=3$, then $\sum_{\text{cyc}}\frac{x^2}{2y^2-y+3}\ge\frac{3}{4}$ 
Let $x,y,z>0$, be such that $x+y+z=3$. Show  that
  $$\dfrac{x^2}{2y^2-y+3}+\dfrac{y^2}{2z^2-z+3}+\dfrac{z^2}{2x^2-x+3}\ge\dfrac{3}{4}.$$

I've tried many things but all have failed.
$$\left(\sum_{\text{cyc}}\dfrac{x^2}{2y^2-y+3}\right)\left(\sum_{\text{cyc}}(2y^2-y+3)\right)\ge (x+y+z)^2=9.$$
But
$$\sum_{\text{cyc}}(2y^2-y+3)=2\sum_{\text{cyc}}x^2+6\ge 12.$$
 A: By C-S $\sum\limits_{cyc}\frac{x^2}{2y^2-y+3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}$. Thus, it remains to prove that $\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2-x^2y+3x^2)}\geq\frac{3}{4}$, which is $\sum\limits_{cyc}(3x^4-x^3y-2x^3z+x^2y^2-x^2yz)\geq0$, which is obvious.  
A: I would go into homogeneous coordinates,
$$A=3=x+y+z$$
$$X=2x-y-z \quad x=(A+X)/3$$
and cyclically for $Y$ and $Z$. The coordinates $X$, $Y$, $Z$ are orthogonal to $A$, so whatever their values, they respect the constraint to the plane. Additionally, $X+Y+Z=0$.
With this, the sum is rewritten:
$$\sum \frac{(A+X)^2}{2(A+Y)^2-3(A+Y)+27}=\sum\frac{9+6X+X^2}{36+3Y+2Y^2}$$
The $X=Y=Z=0$ lies on the symmetry axis, and there, the sum is exactly $\frac{3}{4}$. We now just have to prove that the function is increasing off-axis.
The denominator has no poles on the domain, so the Taylor expansion converges. Just write
$$\frac{1}{4}\sum (1+X/2+X^2/9)(1-(Y/12+Y^2/18)+(Y/12+Y^2/18)^2-(Y/12+Y^2/18)^3+\cdots)$$
and you can extract the second derivatives quite easily.
It may be a bit tricker to prove the function does not turn around and reach below the axis value, but at least you know where to start.
