已知a^2+b^2+c^2=1，求(a-b)(b-c)(c-a)的最小值

.·.·.

mathematica 发表于 2020-9-7 20:00
为什么u v都是非负数呢？有可能是都是负数呀

这是其中一种情况
另一种情况，最小值是0已经求出来了

mathematica

.·.·. 发表于 2020-9-7 21:29
这是其中一种情况
另一种情况，最小值是0已经求出来了

我自己明白了，这题最大值最小值互为相反数，调换一下变量顺序，就能得到最大值，最大值调换变量顺序就能得到最大值

mathematica

mathematica 发表于 2020-9-8 10:06
我自己明白了，这题最大值最小值互为相反数，调换一下变量顺序，就能得到最大值，最大值调换变量顺序就能 ...

拉格朗日乘子法！

1. Clear["Global`*"];
   
2. f=(a-b)*(b-c)*(c-a)+x*(a^2+b^2+c^2-1);
   
3. ans=FullSimplify@Solve[D[f,{{a,b,c,x}}]==0,{a,b,c,x}];
   
4. Grid[ans]
   
5. f/.ans
   

复制代码

\[\begin{array}{cccc}
a\to 0 & b\to \frac{1}{\sqrt{2}} & c\to -\frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} \\
a\to 0 & b\to -\frac{1}{\sqrt{2}} & c\to \frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} \\
a\to -\frac{1}{\sqrt{2}} & b\to \frac{1}{\sqrt{2}} & c\to 0 & x\to \frac{3}{2 \sqrt{2}} \\
a\to -\frac{1}{\sqrt{2}} & b\to 0 & c\to \frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} \\
a\to \frac{1}{\sqrt{2}} & b\to -\frac{1}{\sqrt{2}} & c\to 0 & x\to -\frac{3}{2 \sqrt{2}} \\
a\to \frac{1}{\sqrt{2}} & b\to 0 & c\to -\frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} \\
a\to -\frac{1}{\sqrt{3}} & b\to -\frac{1}{\sqrt{3}} & c\to -\frac{1}{\sqrt{3}} & x\to 0 \\
a\to \frac{1}{\sqrt{3}} & b\to \frac{1}{\sqrt{3}} & c\to \frac{1}{\sqrt{3}} & x\to 0 \\
\end{array}\]

最值
\[\left\{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0\right\}\]

mathematica

abc的排列一共有六种，其中abc、bca、cab，分别带入f(a,b,c)=(a-b)(b-c)(c-a)，得到的是f=(a-b)(b-c)(c-a)，
另外三种排列acb、cba、bac, 分别带入f(a,b,c)=(a-b)(b-c)(c-a), 得到的是-f=-(a-b)(b-c)(c-a)，
假设f=(a-b)(b-c)(c-a)取最大值时，a=x,b=y,c=z,调整xyz的顺序，然后带入f，得到的要么是最大值fmax（很显然fmax大于等于零），要么是最小值fmin（很显然fmin小于等于零），
如果有一个gmin比fmin还小，那么调整取最值时变量的值，就可以得到一个比fmax更大的值，因此矛盾！因此不存在比fmin更小的gmin。
因此求最大值fmax就是求最小值fmin，求最小值就是求最大值，两者互为相反数。
求fmax与fmin就是求h=|(a-b)(b-c)(c-a)|的最大值，h取最大值时，调整变量的顺序，最大值不变，因此可以假设变量a>=b>=c
a-b=u,b-c=v,则a-c=u+v，此时h=u*v*(u+v),


我添加了一些详细的说明，这样mathe的更容易明白！

mathematica

1. Clear["Global`*"];
   
2. Minimize[{(a-b)*(b-c)*(c-a),a^2+b^2+c^2==1},{a,b,c}]
   
3. Minimize[{(a-b)*(b-c)*(c-a),a^2+b^2+c^2==1&&a>=0&&b>=0&&c>=0},{a,b,c}]//FullSimplify
   

复制代码

两个最小值依次是：
\[\left\{-\frac1{\sqrt2},\left\{a\to 0,b\to -\frac1{\sqrt2},c\to \frac1{\sqrt2}\right\}\right\}\]
\[\left\{-\frac1{3 \sqrt3},\left\{a\to 0,b\to\frac{\sqrt5-1}{2\sqrt3} ,c\to \frac{\sqrt5+1}{2\sqrt3}\right\}\right\}\]

mathematica

@.·.·. 按照你的思路，加上我的理解，我重新整理了一个详细的过程。

mathematica

.·.·. 发表于 2020-9-7 15:55
第一问，只需要注意到，在设$a\le b\le c$时，$(a-b)(b-c)\le\frac{(a-c)^2}4$而$|a-c|$在$|a|=|c|=\sq ...

其实你们说的都不对，应该是求最大值，而不是最小值，我在后面说明白了！

王守恩

利用恒等式：$$1=(\cos x)^2+(\sin x\sin y)^2+(\sin x\cos y)^2$$
可设 $a=cos x,b=sinx sin y,c=sin x cos y$ 代入得$$(a-b)(b-c)(c-a)=(\cos x-\sin x\sin y)(\sin x\sin y-\sin x\cos y)(\sin x\cos y-\cos x)\tag{1}
$$观察(1)可得\(\ x=\frac{\pi}{2}\)时，(a-b)(b-c)(c-a)取得最小值$$(0-\sin y)(\sin y-\cos y)(\cos y-0)=\cos y\sin y(\cos y-\sin y)\tag{2}
$$或者\(\ y=0\)时， (a-b)(b-c)(c-a)取得最小值$$(\cos x-0)(0-\sin x)(\sin x-\cos x)=\cos x\sin x(\cos x-\sin x)\tag{3}$$
观察(1+)可得\(\ x=\frac{\pi}{2}\)j时， (a-b)(b-c)(c-a)取得最小值$$(0-\cos y)(\cos y-\sin y)(\sin y-0)=\sin y\cos y(\sin y-\cos y)\tag{4}$$
或者\(\ y=0\)时， (a-b)(b-c)(c-a)取得最小值$$(\cos x-\sin x)(\sin x-0)(0-\cos x)=\sin x\cos x(\sin x-\cos x)\tag{5}$$
合并(2),(3),(4),(5)得，(a-b)(b-c)(c-a)的最小值为$$xy(x-y),x^2+y^2=1$$
也可以这样，(a-b)(b-c)(c-a)的最小值为$$\sqrt{1 - y^2}y(\sqrt{1 - y^2}-y)$$
答案1，当\(y=-\sqrt{2}/2\)时，(a-b)(b-c)(c-a)的最小值为$$-\sqrt{2}/2$$
答案2，当\(y=\frac{\sqrt{5}-1}{\sqrt{12}}\)时，(a-b)(b-c)(c-a)的最小值为$$-\sqrt{3}/9$$
答案3，当\(y=\frac{\sqrt{5}+1}{\sqrt{12}}\)时，(a-b)(b-c)(c-a)的最大值为$$\sqrt{3}/9$$

王守恩

王守恩 发表于 2020-11-30 09:32
利用恒等式：$$1=(\cos x)^2+(\sin x\sin y)^2+(\sin x\cos y)^2$$
可设 $a=cos x,b=sinx sin y,c=sin x c ...

\(1=(\cos a)^2+(\sin a)^2\)
\(1=(\cos a)^2+(\sin a\cos b)^2+(\sin a\sin b)^2\)
\(1=(\cos a)^2+(\sin a\cos b)^2+(\sin a\sin b\cos c)^2+(\sin a\sin b\sin c)^2\)
\(1=(\cos a)^2+(\sin a\cos b)^2+(\sin a\sin b\cos c)^2+(\sin a\sin b\sin c\cos d)^2+(\sin a\sin b\sin c\sin d)^2\)

mathematica

王守恩 发表于 2020-11-30 19:37
\(1=(\cos a)^2+(\sin a)^2\)
\(1=(\cos a)^2+(\sin a\cos b)^2+(\sin a\sin b)^2\)
\(1=(\cos a)^2+(\ ...

利用kkt条件，先求出所有可能的极值点

1. Clear["Global`*"];
   
2. f=(a-b)*(b-c)*(c-a)+x*(a^2+b^2+c^2-1)-x1*a-x2*b-x3*c;
   
3. ans=ToRadicals@FullSimplify@Solve[D[f,{{a,b,c,x}}]==0&&x1*a==0&&x2*b==0&&x3*c==0,{a,b,c,x,x1,x2,x3}];
   
4. aaa=Append[#,FullSimplify[(f/.#)]]&/@ans;(*增加一列函数值*)
   
5. bbb=Sort[aaa,#1[[8]]>#2[[8]]&];(*按照函数值降序排列*)
   
6. Grid[bbb,Alignment->Left]
   
7. (*选择abc全部大于等于零的情况*)
   
8. ccc=Select[bbb,(a/.#[[1]])>=0&&(b/.#[[2]])>=0&&(c/.#[[3]])>=0&];
   
9. Grid[ccc,Alignment->Left]
   

复制代码

所有可能的极值点

\[\begin{array}{llllllll}
a\to \frac{1}{\sqrt{2}} & b\to -\frac{1}{\sqrt{2}} & c\to 0 & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to -\frac{1}{\sqrt{2}} & b\to 0 & c\to \frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to \frac{1}{\sqrt{2}} & b\to -\frac{1}{\sqrt{2}} & c\to 0 & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to -\frac{1}{\sqrt{2}} & b\to 0 & c\to \frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to 0 & b\to \frac{1}{\sqrt{2}} & c\to -\frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to 0 & b\to \frac{1}{\sqrt{2}} & c\to -\frac{1}{\sqrt{2}} & x\to -\frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{\sqrt{2}} \\
a\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to 0 & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to \frac{\sqrt{5}}{3} & \frac{1}{3 \sqrt{3}} \\
a\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to 0 & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to -\frac{\sqrt{5}}{3} & \frac{1}{3 \sqrt{3}} \\
a\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to 0 & c\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to -\frac{\sqrt{5}}{3} & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to 0 & c\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to \frac{\sqrt{5}}{3} & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to \frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to -\frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to 1 & b\to 0 & c\to 0 & x\to 0 & \text{x1}\to 0 & \text{x2}\to -1 & \text{x3}\to 1 & 0 \\
a\to -1 & b\to 0 & c\to 0 & x\to 0 & \text{x1}\to 0 & \text{x2}\to -1 & \text{x3}\to 1 & 0 \\
a\to \frac{1}{\sqrt{3}} & b\to \frac{1}{\sqrt{3}} & c\to \frac{1}{\sqrt{3}} & x\to 0 & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & 0 \\
a\to -\frac{1}{\sqrt{3}} & b\to -\frac{1}{\sqrt{3}} & c\to -\frac{1}{\sqrt{3}} & x\to 0 & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & 0 \\
a\to 0 & b\to 1 & c\to 0 & x\to 0 & \text{x1}\to 1 & \text{x2}\to 0 & \text{x3}\to -1 & 0 \\
a\to 0 & b\to -1 & c\to 0 & x\to 0 & \text{x1}\to 1 & \text{x2}\to 0 & \text{x3}\to -1 & 0 \\
a\to 0 & b\to 0 & c\to 1 & x\to 0 & \text{x1}\to -1 & \text{x2}\to 1 & \text{x3}\to 0 & 0 \\
a\to 0 & b\to 0 & c\to -1 & x\to 0 & \text{x1}\to -1 & \text{x2}\to 1 & \text{x3}\to 0 & 0 \\
a\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to 0 & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to \frac{\sqrt{5}}{3} & -\frac{1}{3 \sqrt{3}} \\
a\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to 0 & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to -\frac{\sqrt{5}}{3} & -\frac{1}{3 \sqrt{3}} \\
a\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to 0 & c\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to -\frac{\sqrt{5}}{3} & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
a\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to 0 & c\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to \frac{\sqrt{5}}{3} & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to \frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to -\sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to -\sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to -\frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
a\to -\frac{1}{\sqrt{2}} & b\to \frac{1}{\sqrt{2}} & c\to 0 & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
a\to \frac{1}{\sqrt{2}} & b\to 0 & c\to -\frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
a\to \frac{1}{\sqrt{2}} & b\to 0 & c\to -\frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
a\to -\frac{1}{\sqrt{2}} & b\to \frac{1}{\sqrt{2}} & c\to 0 & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
a\to 0 & b\to -\frac{1}{\sqrt{2}} & c\to \frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
a\to 0 & b\to -\frac{1}{\sqrt{2}} & c\to \frac{1}{\sqrt{2}} & x\to \frac{3}{2 \sqrt{2}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{\sqrt{2}} \\
\end{array}\]

所有非负的极值点
\[\begin{array}{llllllll}
a\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to 0 & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to -\frac{\sqrt{5}}{3} & \frac{1}{3 \sqrt{3}} \\
a\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to 0 & c\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to -\frac{\sqrt{5}}{3} & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to -\frac{1}{2 \sqrt{3}} & \text{x1}\to -\frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & \frac{1}{3 \sqrt{3}} \\
a\to 1 & b\to 0 & c\to 0 & x\to 0 & \text{x1}\to 0 & \text{x2}\to -1 & \text{x3}\to 1 & 0 \\
a\to \frac{1}{\sqrt{3}} & b\to \frac{1}{\sqrt{3}} & c\to \frac{1}{\sqrt{3}} & x\to 0 & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to 0 & 0 \\
a\to 0 & b\to 1 & c\to 0 & x\to 0 & \text{x1}\to 1 & \text{x2}\to 0 & \text{x3}\to -1 & 0 \\
a\to 0 & b\to 0 & c\to 1 & x\to 0 & \text{x1}\to -1 & \text{x2}\to 1 & \text{x3}\to 0 & 0 \\
a\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & b\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & c\to 0 & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to 0 & \text{x3}\to \frac{\sqrt{5}}{3} & -\frac{1}{3 \sqrt{3}} \\
a\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & b\to 0 & c\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to 0 & \text{x2}\to \frac{\sqrt{5}}{3} & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
a\to 0 & b\to \sqrt{\frac{1}{6} \left(\sqrt{5}+3\right)} & c\to \sqrt{\frac{1}{6} \left(3-\sqrt{5}\right)} & x\to \frac{1}{2 \sqrt{3}} & \text{x1}\to \frac{\sqrt{5}}{3} & \text{x2}\to 0 & \text{x3}\to 0 & -\frac{1}{3 \sqrt{3}} \\
\end{array}\]