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ILoveBainiu 写了： 2025年 6月 26日 23:14 The expression to minimize is sqrt(x² + 1²) + sqrt(y² + 3²) + sqrt(z² + 4²).

Each term in the sum can be interpreted as the length of the hypotenuse of a right triangle, or as the distance between two points in a 2D plane.

Let's visualize this as the length of a path made of three segments:

A segment with a horizontal displacement of x and a vertical displacement of 1. Its length is sqrt(x² + 1²).
A segment with a horizontal displacement of y and a vertical displacement of 3. Its length is sqrt(y² + 3²).
A segment with a horizontal displacement of z and a vertical displacement of 4. Its length is sqrt(z² + 4²).
The total length of this path is the sum we want to minimize. The path starts at an origin point, say (0,0), and ends at the point whose coordinates are the sum of the displacements.

The total horizontal displacement is x + y + z.
The total vertical displacement is 1 + 3 + 4 = 8.
We are given the constraint that x + y + z = 6. Therefore, the path connects the starting point (0, 0) to the fixed endpoint (6, 8).

The sum of the lengths of the path segments is minimized when the path is a straight line. The minimum value of the expression is therefore the straight-line distance between the point (0, 0) and the point (6, 8).

We can calculate this distance using the distance formula:

Distance = sqrt((6 - 0)² + (8 - 0)²)
Distance = sqrt(6² + 8²)
Distance = sqrt(36 + 64)
Distance = sqrt(100)
Distance = 10

Thus, the smallest possible value of the expression is 10.

看看gemini多牛逼

yuzhou 写了： 2025年 6月 26日 23:34 sqrt(x^2+1) + sqrt(y^2+9) + sqrt(z^2+16) >=sqrt(2x) + sqrt(6y)+sqrt(8z)

yuzhou 写了： 2025年 6月 26日 23:34 sqrt(x^2+1) + sqrt(y^2+9) + sqrt(z^2+16) >=sqrt(2x) + sqrt(6y)+sqrt(8z)

有题目对称性得到

3（sqrt(x^2+1) + sqrt(y^2+9) + sqrt(z^2+16) ） >= (sqrt(2)+sqrt(6)+sqrt(8)) (sqrt(x) + sqrt(y)+sqrt(z))


(sqrt(x) + sqrt(y)+sqrt(z))^2 =x+y+z + sqrt(x) sqrt(y)+ sqrt(y) sqrt(z)+ sqrt(y) sqrt(z) =6+ sqrt(x) sqrt(y)+ sqrt(y) sqrt(z)+ sqrt(x) sqrt(z)


由柯西不等式 （x+y+x)^2>=(sqrt(x) sqrt(y)+ sqrt(y) sqrt(z)+ sqrt(x) sqrt(z))^2
sqrt(x) sqrt(y)+ sqrt(y) sqrt(z)+ sqrt(x) sqrt(z)<=6

3倍的最小值 是(sqrt(2)+sqrt(6)+sqrt(8))*(sqrt(12))/3
最小值是 （2sqrt(6) +6sqrt(2)+8sqrt(3)）/3

ILoveBainiu 写了： 2025年 6月 26日 23:14 The expression to minimize is sqrt(x² + 1²) + sqrt(y² + 3²) + sqrt(z² + 4²).

Each term in the sum can be interpreted as the length of the hypotenuse of a right triangle, or as the distance between two points in a 2D plane.

Let's visualize this as the length of a path made of three segments:

A segment with a horizontal displacement of x and a vertical displacement of 1. Its length is sqrt(x² + 1²).
A segment with a horizontal displacement of y and a vertical displacement of 3. Its length is sqrt(y² + 3²).
A segment with a horizontal displacement of z and a vertical displacement of 4. Its length is sqrt(z² + 4²).
The total length of this path is the sum we want to minimize. The path starts at an origin point, say (0,0), and ends at the point whose coordinates are the sum of the displacements.

The total horizontal displacement is x + y + z.
The total vertical displacement is 1 + 3 + 4 = 8.
We are given the constraint that x + y + z = 6. Therefore, the path connects the starting point (0, 0) to the fixed endpoint (6, 8).

The sum of the lengths of the path segments is minimized when the path is a straight line. The minimum value of the expression is therefore the straight-line distance between the point (0, 0) and the point (6, 8).

We can calculate this distance using the distance formula:

Distance = sqrt((6 - 0)² + (8 - 0)²)
Distance = sqrt(6² + 8²)
Distance = sqrt(36 + 64)
Distance = sqrt(100)
Distance = 10

Thus, the smallest possible value of the expression is 10.

看看gemini多牛逼

Yesterday 写了： 2025年 6月 27日 00:52 看来你得到白牛指导了。用几何解决代数问题是另一个维度

ILoveBainiu 写了： 2025年 7月 1日 03:39 菌斑僵菌很多看不懂拉格朗日，就不贴了

bam 写了： 2025年 7月 1日 03:43 这不虾扯蛋吗，但凡不是文科生谁不懂拉格朗日乘数法啊

ILoveBainiu 写了： 2025年 7月 1日 03:54 千老未必会

lpz9678 写了： 2025年 6月 26日 22:28

lpz9678 写了： 2025年 6月 26日 22:28

SidRat 写了： 2025年 7月 1日 08:08 数形结合，答案是10啦！
如下图所示，要使3项之和最小，那么就需要A、I、G、F四点共线，此时AF = sqrt(6^2 + (4+4)^2) = 10。

SidRat 写了： 2025年 7月 1日 08:08 数形结合，答案是10啦！
如下图所示，要使3项之和最小，那么就需要A、I、G、F四点共线，此时AF = sqrt(6^2 + (4+4)^2) = 10。

LiuQiangDong 写了： 2025年 7月 1日 05:07 俩子空间相切吧
gemini的解法是做题家解法
投机取巧
好处是速度快
容易拿分
坏处是不普世
限制思维扩展