lean_workbook_plus_11047

For $a+b+c>=\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ , we have\n\n $(a+b+c)^2 >= (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) >= 9$ .\n\nAnd with the Power--Mean Inequality, we get\n\n $(\frac{a^3+b^3+c^3}{3})^{\frac{1}{3}}>=\frac{a+b+c}{3}$ ,\n\nThe above inequality is equivalent to\n\n $a^3+b^3+c^3>=\frac{(a+b+c)^3}{9}$ .\n\nIn order to prove the origin inequality, we only need prove the inequality as follows.\n\n $\frac{(a+b+c)^3}{9}>=a+b+c$ \n\nThe inequality is exactly the inequality\n\n $(a+b+c)^2 >= 9$ .\n\nThus, the proof of the origin inequality is completed.