如何获得Sqrt [3x + 1] + Sqrt [3y + 1] + Sqrt [3z + 1]的最小值？
让
f[x_,y_,z_] := Sqrt[3x+1]+Sqrt[3y+1]+Sqrt[3z+1]
Since 0<=x<=1,0<=y<=1,0<=z<=1, we have
0<=x^2<=x,0<=y^2<=y,0<=z^2<=z.
Hence,
3a+1 >= a^2 + 2a + 1 = (a+1)^2, where a in {x,y,z}.
Consequently,
f[x,y,z] >= x+1+y+1+z+1 = 4,
Where the equality holds if and only if (x==0&&y==0||z==1)||...
Minimize[{f[x,y,z],x>=0&&y>=0&&z>=0&&x+y+z==1},{x,y,z}]
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芭隘的盘石
In[1]:= f[x_,y_,z_]:=Sqrt[3x+1]+Sqrt[3y+1]+Sqrt[3z+1] In[2]:= Minimize[{f[x,y,z],x>=0,y>=0,z>=0,x+y+z==1},{x,y,z}] Out[2]= {4,{x->1,y->0,z->0}}In[3]:= Thread[D[f[x,y,z] - [Lambda](x+y+z-1), {{x,y,z,[Lambda]}}]==0]; Reduce[Join[%,{x>=0,y>=0,z>=0}],{x,y,z,[Lambda]},Reals] {f[x,y,z],D[f[x, y, z], {{x, y, z}, 2}]}/.ToRules[%] Out[4]= x==1/3&&y==1/3&&z==1/3&&[Lambda]==3/(2 Sqrt[2]) Out[5]= {3 Sqrt[2],{{-(9/(8 Sqrt[2])),0,0},{0,-(9/(8 Sqrt[2])),0},{0,0,-(9/(8 Sqrt[2]))}}}坛沤疲撑拆
f[x_, y_, z_] := Sqrt[3 x + 1] + Sqrt[3 y + 1] + Sqrt[3 z + 1] g1 = ContourPlot3D[f[x, y, z] == 4, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, AxesLabel -> {x,y,z}, MeshFunctions -> {#3 &}, ContourStyle -> {Blue, Opacity[0.5]}]; g2 = ContourPlot3D[ x + y + z == 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, AxesLabel -> {x,y,z}, MeshFunctions -> {#2 &}, ContourStyle -> {Green, Opacity[0.5]}]; Show[g1, g2, Graphics3D[{PointSize[0.05], Red, Point[{1, 0, 0}]}], ViewPoint -> {1.1`, -2.4`, 1.7`}]坍锭嘉韭蓝