lean_workbook_plus_61538

Since \n\n $R^2=\frac{a^2b^2c^2}{16S^2}=\frac {a^2b^2c^2}{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$ , \nusing the substitution $a^2=\frac{y+z}{2}$ etc., the inequality \n\n $\displaystyle \frac{a^4+b^4+c^4}{a^2+b^2+c^2} \ge 3R^2$ \n\nbecomes as follows \n\n $xy(x-y)^2+yz(y-z)^2+zx(z-x)^2 \ge 0.$ \n