SOLUTION: The real numbers x_1, x_2, x_3, x_4, and x_5 satisfy x_1 + x_2 + x_3 + x_4 + x_5 = 8, x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 12, x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3 = 16.

Algebra ->  Expressions-with-variables -> SOLUTION: The real numbers x_1, x_2, x_3, x_4, and x_5 satisfy x_1 + x_2 + x_3 + x_4 + x_5 = 8, x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 12, x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3 = 16.       Log On


   



Question 1209927: The real numbers x_1, x_2, x_3, x_4, and x_5 satisfy
x_1 + x_2 + x_3 + x_4 + x_5 = 8,
x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 12,
x_1^3 + x_2^3 + x_3^3 + x_4^3 + x_5^3 = 16.

Let m be the smallest possible value of x_5, and let M be the largest possible value of x_5. Enter the ordered pair (m,M).

Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let $S_1 = \sum_{i=1}^5 x_i = 8$, $S_2 = \sum_{i=1}^5 x_i^2 = 12$, and $S_3 = \sum_{i=1}^5 x_i^3 = 16$.
Let $y_i = x_i$ for $i=1,2,3,4$.
Then $y_1 + y_2 + y_3 + y_4 = 8 - x_5$, $y_1^2 + y_2^2 + y_3^2 + y_4^2 = 12 - x_5^2$, and $y_1^3 + y_2^3 + y_3^3 + y_4^3 = 16 - x_5^3$.
By the Cauchy-Schwarz inequality,
$(y_1 + y_2 + y_3 + y_4)^2 \le 4(y_1^2 + y_2^2 + y_3^2 + y_4^2)$.
Substituting the given equations, we have
$(8 - x_5)^2 \le 4(12 - x_5^2)$.
$64 - 16x_5 + x_5^2 \le 48 - 4x_5^2$.
$5x_5^2 - 16x_5 + 16 \le 0$.
Let $f(x_5) = 5x_5^2 - 16x_5 + 16$.
We find the roots of $f(x_5) = 0$ using the quadratic formula:
$x_5 = \frac{16 \pm \sqrt{16^2 - 4(5)(16)}}{2(5)} = \frac{16 \pm \sqrt{256 - 320}}{10} = \frac{16 \pm \sqrt{-64}}{10}$.
Since the discriminant is negative, there are no real roots. Thus, $f(x_5) > 0$ for all real $x_5$.
However, we can apply the power mean inequality.
Let $y_1, y_2, y_3, y_4$ be real numbers.
Let $A = \frac{1}{4} \sum_{i=1}^4 y_i$, $Q = \sqrt{\frac{1}{4} \sum_{i=1}^4 y_i^2}$, and $C = \sqrt[3]{\frac{1}{4} \sum_{i=1}^4 y_i^3}$.
$A \le Q \le C$.
Thus $A^2 \le Q^2$ and $Q^3 \le C^3$.
$A = \frac{8-x_5}{4}$, $Q^2 = \frac{12-x_5^2}{4}$, $C^3 = \frac{16-x_5^3}{4}$.
$A^2 \le Q^2$:
$(\frac{8-x_5}{4})^2 \le \frac{12-x_5^2}{4}$
$\frac{(8-x_5)^2}{16} \le \frac{12-x_5^2}{4}$
$(8-x_5)^2 \le 4(12-x_5^2)$
$64 - 16x_5 + x_5^2 \le 48 - 4x_5^2$
$5x_5^2 - 16x_5 + 16 \le 0$
$Q^3 \le C^3$:
$(\frac{12-x_5^2}{4})^{3/2} \le (\frac{16-x_5^3}{4})$
$(\frac{12-x_5^2}{4})^3 \le (\frac{16-x_5^3}{4})^2$
$(12-x_5^2)^3 \le 16(16-x_5^3)^2$
Consider the case where $y_1=y_2=y_3=y_4$.
Then $4y_1 = 8-x_5$, $4y_1^2 = 12-x_5^2$, $4y_1^3 = 16-x_5^3$.
$y_1 = \frac{8-x_5}{4}$.
$4(\frac{8-x_5}{4})^2 = 12-x_5^2$
$\frac{(8-x_5)^2}{4} = 12-x_5^2$
$(8-x_5)^2 = 4(12-x_5^2)$
$64 - 16x_5 + x_5^2 = 48 - 4x_5^2$
$5x_5^2 - 16x_5 + 16 = 0$
As shown before, this equation has no real roots.
$4(\frac{8-x_5}{4})^3 = 16-x_5^3$
$\frac{(8-x_5)^3}{16} = 16-x_5^3$
$(8-x_5)^3 = 16(16-x_5^3)$
$512 - 192x_5 + 24x_5^2 - x_5^3 = 256 - 16x_5^3$
$15x_5^3 + 24x_5^2 - 192x_5 + 256 = 0$
Let $g(x_5) = 15x_5^3 + 24x_5^2 - 192x_5 + 256$.
$g(0) = 256$, $g(1) = 15+24-192+256 = 103$, $g(2) = 15(8)+24(4)-192(2)+256=120+96-384+256 = 88$, $g(3) = 15(27)+24(9)-192(3)+256=405+216-576+256=301$.
$g(4)=15(64)+24(16)-192(4)+256=960+384-768+256=832$
$g(-1) = -15+24+192+256>0$.
$g(-2) = -15(8)+24(4)+192(2)+256=-120+96+384+256=616$
We require $5x_5^2-16x_5+16 \le 0$ to hold.
$x_5 = \frac{16\pm \sqrt{16^2-4(5)(16)}}{10}$.
$x_5$ is complex.
Final Answer: The final answer is $\boxed{(m,M)}$