$$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$

$$f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}$$

$$ = {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$$

$$g(x) = {e^{ - x}} - 2{e^x}$$

$$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$$

$$\Rightarrow$$ f(x) is increasing and g(x) is decreasing function.

$$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)$$

$$ \Rightarrow {{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$$

$$ = {\alpha ^2} - 5\alpha + 6 < 0$$

$$ = (\alpha - 2)(\alpha - 3) < 0$$

$$ = \alpha \in (2,\,3)$$