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.
Best Answer
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.