Because $x^{2}+1 \geq 2x$, we have
$$\frac{\left(x^{2}+1\right)^{7}}{2^{7}} \geq \frac{(2x)^{7}}{2^{7}} = x^{7}, \quad \text{and} \quad \frac{x^{2}+1}{2} \geq x,$$
and deduce
$\left(x^{2}+1\right) \cdot \left(\frac{\left(x^{2}+1\right)^{6}}{2^{7}}+\frac{1}{2}\right) = \frac{\left(x^{2}+1\right)^{7}}{2^{7}} + \frac{x^{2}+1}{2} \geq x^{7} + x = x\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$.
Therefore
$$\frac{\left(x^{2}+1\right)^{6}}{2^{7}}+\frac{1}{2} \geq x^{5}-x^{3}+x$$
and the equality holds when $x=1$.