lean_workbook_plus_12156

Using the inequality which you proved in the first place, we have $2(a^2+b^2+c^2)\geq 2(ab+bc+ca)$ , yielding \n \n $2(a^2+b^2+c^2)+2(ab+bc+ca)\geq 4(ab+bc+ca)$ , but taking the reciprocal of both sides for positive real numbers $a,\ b,\ c$ , \n \n we have $\frac{1}{2(a^2+b^2+c^2)+2(ab+bc+ca)}\leq \frac{1}{4(ab+bc+ac)}.$