To solve this inequality, we can take the natural logarithm of both sides to simplify the expression:

[ \ln\left((x^{2} + x + 1)^{x^2 - 5x + 6}\right) > \ln\left((x+2)^{x^2 - 5x + 6}\right) ]

Using the properties of logarithms, we can simplify this to:

[ (x^2 - 5x + 6)\ln(x^{2} + x + 1) > (x^2 - 5x + 6)\ln(x+2) ]

Since the inequality holds for all real values of x, we can divide both sides by the common factor of (x^2 - 5x + 6):

[ \ln(x^{2} + x + 1) > \ln(x+2) ]

Now, we can drop the natural logarithm function by exponentiating both sides:

[ x^{2} + x + 1 > x + 2 ]

This simplifies to:

[ x^2 + x - 1 > 0 ]

Now, we need to solve this quadratic inequality to find the range of values for x that satisfy the original inequality:

The roots of the quadratic equation (x^2 + x - 1 = 0) can be found using the quadratic formula:

[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

where a = 1, b = 1, and c = -1:

[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} ]

So, the roots are (x = \frac{-1 + \sqrt{5}}{2}) and (x = \frac{-1 - \sqrt{5}}{2}).

Therefore, the solution to the inequality is:

[ x \in \left(-\infty, \frac{-1 - \sqrt{5}}{2}\right) \cup \left(\frac{-1 + \sqrt{5}}{2}, +\infty\right) ]