Divide each term of $a^{2}+1$ by $2$ to get $\frac{1}{2}a^{2}+\frac{1}{2}$.
$$\frac{1}{2}a^{2}+\frac{1}{2}\geq a$$
Subtract $a$ from both sides.
$$\frac{1}{2}a^{2}+\frac{1}{2}-a\geq 0$$
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$\frac{1}{2}a^{2}+\frac{1}{2}-a=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $\frac{1}{2}$ for $a$, $-1$ for $b$, and $\frac{1}{2}$ for $c$ in the quadratic formula.
