lean_workbook_plus_59690

The first inequality is equivalent to: $\sqrt{\frac{2}{3}+\frac{abc}{a^3+b^3+c^3}}\ge\sqrt{\frac{bc+ca+ab}{a^2+b^2+c^2}}\Leftrightarrow \frac{1}{3}-\frac{abc}{a^3+b^3+c^3}\le\frac{a^2+b^2+c^2-ab-bc-ca}{a^2+b^2+c^2}\Leftrightarrow \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{3(a^3+b^3+c^3)}\le\frac{a^2+b^2+c^2-ab-bc-ca}{a^2+b^2+c^2}\Leftrightarrow (a^2+b^2+c^2-ab-bc-ca)(\frac{1}{a^2+b^2+c^2}-\frac{a+b+c}{3(a^3+b^3+c^3)})\ge 0$