lean_workbook_plus_78790

Given positive numbers $a$, $b$, and $c$, prove that $\frac{4}{a^{2}+bc}\le\frac{1}{a^{2}}+\frac{1}{bc}$, which implies $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$.