lean_workbook_plus_3407

Let $a + b + c = p$ and $ab + bc + ca = q$ \n\n $8(p^2 - 2q)(q) \le p^4$ \n\n $16q^2 - 8p^2q + p^4 \ge 0$ \n\n $(4q - p^2)^2 \ge 0$ \n\nEqualty holds when $4q - p^2 = 0$ <--> $4ab + 4ac + 4bc = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$ \n\n $2ab + 2ac + 2bc = a^2 + b^2 + c^2$ \n\n $a^2 - 2(b + c)a + (b - c)^2 = 0$ \n\nThe discriminant is $16bc$ and must be $\ge 0$ for there to be an $a$ that satisfies the solution. Similarly, transposing the variables cyclically, we also get that $bc \ge 0$ and $ca \ge 0$ So they either are all $\ge 0$ or all $\le 0$ . \n\nIf they are all greater than 0, we have $a = \frac {2(b + c)\pm4\sqrt {bc}}{2} = (\sqrt {b}\pm\sqrt {c})^2$ \n\nso the solutions to the equality are $((\sqrt {b}\pm\sqrt {c})^2,b,c)$ with $b, c\ge0$ when a,b,c>0. \n\nSimilarly,we find that if they are all negative, then $( - (\sqrt {x}\pm\sqrt {y})^2, - x, - y)$ with $x,y \ge 0$ constitutes all other solutions (when $a,b,c \le 0$ ).