lean_workbook_plus_52659

ab+bc+ca=1\Longleftrightarrow a(b+c)=1-bc$etc, \n \n the given inequality can be rewritten as$ \frac{1-bc}{b+c}+\frac{1-ca}{c+a}+\frac{1-ab}{a+b}\geq \sqrt{3}\Longleftrightarrow a+b+c\geq \sqrt{3}$, \n \n which is true from$(a+b+c)^2\geq 3\Longleftrightarrow (a+b+c)^2\geq 3(ab+bc+ca)\Longleftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0.$