Let $f:R \to R$ and $g:R \to R$ be two functions defined by $f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$ and $g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}$. Then, for which of the following range of $\alpha$, the inequality $$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right)$$ holds ?

<p>$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$</p> <p>$f'(x) = {{2x} \over {{x^2} + 1}} + {e^{ - x}}$</p> <p>$= {2 \over {x + {1 \over x}}} + {e^{ - x}} > 0\,\,\forall x \in R$</p> <p>$g(x) = {e^{ - x}} - 2{e^x}$</p> <p>$g'(x) - - {e^{ - x}} - 2{e^x} < 0\,\,\,\,\forall x \in R$</p> <p>$\Rightarrow$ f(x) is increasing and g(x) is decreasing function.</p> <p>$$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha - {5 \over 3}} \right)} \right)$$</p> <p>$\Rightarrow {{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$</p> <p>$= {\alpha ^2} - 5\alpha + 6 < 0$</p> <p>$= (\alpha - 2)(\alpha - 3) < 0$</p> <p>$= \alpha \in (2,\,3)$</p>