已知$a^2+b^2+c^2=1$求$abc(a+b+c)$的最小值.(2018辽宁预赛解答压轴题)
不妨设$a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$,令$u^2=tv^2$要求最小值只需考虑$a,b>0,c<0,a+b+c>0$此时$t<\dfrac{2}{3}$则
$\dfrac{abc(a+b+c)}{(a^2+b^2+c^2)^2}=\dfrac{3uw^3}{(9u^2-6v^2)^2}\ge \dfrac{3u(3uv^2-2u^3-2\sqrt{(u^2-v^2)^3})}{(9u^2-6v^2)^2}$
$=\dfrac{3t-2t^2-2\sqrt{t(t-1)^3}}{3(3t-2)^2}=f(t)$
求导$f^{'}(t)=\dfrac{-t^3+3t-2-\sqrt{(t-1)^3t}(t+6)}{3\sqrt{(t-1)^3t}(3t-2)^3}$
令$-t^3+3t-2=\sqrt{(t-1)^3t}(t+6)$两边平方整理得$(1-t)^3(9t^2+36t+4)=0$此时有根
$t=1,t=-2-\dfrac{4\sqrt{2}}{3},t=-2+\dfrac{4\sqrt{2}}{3}$故$f(t)_{min}=f(-2-\dfrac{4\sqrt{2}}{3})=\dfrac{-1-\sqrt{2}}{16}$
备注:$a,b,c\in R\Leftrightarrow (a-b)^2(b-c)^2(c-a)^2\ge0$
注意到“必背”恒等式$(a-b)^2(b-c)^2(c-a)^2=27(4(u^2-v^2)^3-(w^3-3uv^2+2u^3)^2)\ge0$得
$w^3\in[3uv^2-2u^3-2\sqrt{(u^2-v^2)^3},3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}]$
注:(Schur's ineq)$\sum\limits_{cyc}a(a-b)(a-c)\ge0\Leftrightarrow w^3+3u^3\ge4uv^2$