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I would appreciate if somebody could help me with the following problem:

Q: Find minimum (where $0<x,\:y,\:z<1$) $$\sqrt{x^2+y^2-2y+2}+\sqrt{x^2+z^2-2x+2}+\sqrt{y^2+z^2-2z+2}$$

My work: Just as an expression with two roots can be interpreted as a distance in a plane, $$(x,1,0),\:(0,y,1),\:(1,0,z)$$ the above expression is interpreted as a distance in space to find a value that minimizes the sum of the distances, but it does not work. Please let me know how to solve it algebraically or geometrically. thank you

Young
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    Notice the symmetry of the problem. Also, for any $~0 < k < 1,~$ the expression $~k^2 + (1-k)^2~$ is minimized at $~k = \dfrac{1}{2}.~$ This should informally suggest an optimum value for each variable $~x,y,z.~$ Then, you would have to prove that this informal suggestion is optimal. – user2661923 Apr 25 '23 at 00:56
  • You can still use the Minkowski's inequality to turn this into a single-variable optimization problem which nicely happens to be minimized at the unique triplet the aforementioned Minkowski's application is also an equality.

    Geometrically, this might have a cute solution by some extra constructions of the cube on the some of its faces, like the geometric solution for Fermat point

     – dezdichado Apr 25 '23 at 01:01

1 Answers1

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By the Cauchy–Schwarz inequality, we have:

$$\sqrt {x^2+(1-y)^2+1^2}\sqrt {(\frac{1}{2})^2+(\frac{1}{2})^2+1^2}\geq \frac{x}{2}+\frac{1-y}{2}+1 \\ \implies \sqrt {x^2+(1-y)^2+1^2} \geq \sqrt {\frac{2}{3}}(\frac{1-y+x}{2}+1).$$

Similarly,

$$\sqrt {z^2+(1-x)^2+1^2} \geq \sqrt {\frac{2}{3}}(\frac{1-x+z}{2}+1), \\ \sqrt {y^2+(1-z)^2+1^2} \geq \sqrt {\frac{2}{3}}(\frac{1-z+y}{2}+1).$$

Therefore,

$$\sqrt {x^2+(1-y)^2+1^2}+\sqrt {z^2+(1-x)^2+1^2}+\sqrt {y^2+(1-z)^2+1^2} \geq \\\sqrt {\frac{2}{3}}(\frac{3}{2}+3)=3\sqrt {\frac{3}{2}}. $$

The equality happens at $x=y=z=\frac{1}{2}$.

Reza Rajaei
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