Find the minimum of $x^2+y^2+z^2$ under several constraints

Question

Let $x,y,z \in (-1;1)$ such that $$13\left( {xy + yz + zx + 1} \right) + 14\left( {x + y + z + xyz} \right) = 0.$$ Find the minimum value of the following expression $$P=x^2+y^2+z^2.$$

I guess the minimum value is $\frac{3}{4}$ and I will try to prove that prediction. From the conditions of the variables, I thought of trigonometry. However, it cannot be used due to the conditions $$13\left( {xy + yz + zx + 1} \right) + 14\left( {x + y + z + xyz} \right) = 0.$$ I also thought of the $pqr$ method, but the variables are real numbers (which can be negative), so I couldn't implement the idea. Please provide some suggestions. Thank you.

Lion Heart $13\left( {xy + yz + zx + 1} \right) + 14\left( {x + y + z + xyz} \right) = 0.$ — Math_fun2006, Apr 13 '23 at 09:44
@Math_fun2006 The pqr method is applied to real numbers (including negative ones). So why couldn't you apply the idea? — Calvin Lin, Apr 13 '23 at 13:13
@CalvinLin Simply let $a = x + 1, b = y + 1, c = z + 1$ to avoid negative and use pqr. — River Li, Apr 13 '23 at 14:07
@RiverLi Thank you, this is a great idea. I have performed and obtained $$14abc+2(a+b+c)=ab+bc+ca+4$$ But I don't know how to proceed with $$P=(a−1)^2+(b−1)^2+(c−1)^2$$ . Can you help me? Thank you! — Math_fun2006, Apr 13 '23 at 18:52
@Math_fun2006 Although pqr works, it is not nice. I will give another proof. — River Li, Apr 14 '23 at 00:15

score 5 · Accepted Answer · answered Apr 14 '23 at 00:22

5

Answer: The minimum is $3/4$ when $x = y = z = -1/2$.

Proof:

WLOG, assume that $(y + 1/2)(z + 1/2)\ge 0$ (i.e. $y$ and $z$ are on the same side of $-1/2$; Pigeonhole Principle).
Note: Or note that $(x+1/2)(y+1/2)\cdot (y+1/2)(z+1/2) \cdot (z+1/2)(x+1/2) \ge 0$. So at least one of the three terms is non-nagative. WLOG, assume that $(y + 1/2)(z + 1/2)\ge 0$.

We have \begin{align*} &x^2 + y^2 + z^2 - \frac34 + \frac29\cdot [13(xy + yz + zx + 1) + 14(x + y + z + xyz)]\\[6pt] ={}& \frac{28}{9}(x + 1)(y + 1/2)(z + 1/2) + \frac{1}{36}(6x + 4y + 4z + 7)^2 + \frac59(y-z)^2\\[6pt] \ge{}& 0. \end{align*}

We are done.

answered Apr 14 '23 at 00:22

River Li

49,125

Your solution is great. Thank you very much! I have a question about how to add just enough and combine it logically as mentioned above. – Math_fun2006 Apr 14 '23 at 05:17
You can used the method of unknown coefficients. $L(x, \lambda) = f(x) + \lambda g(x)$. Then $\frac{\partial L}{\partial x} = 0$ at the minimizer. See more examples here: https://math.stackexchange.com/questions/449755/prove-that-a-d2b-c2-geq-1-6-if-a24b2-4-and-cd-4/4539078#4539078 – River Li Apr 14 '23 at 06:04

Find the minimum of $x^2+y^2+z^2$ under several constraints

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