Find all real numbers $x, y,z,t \in [0, \infty)$ so that $$x + y + z \le t, \,\,\, x^2 + y^2 + z^2 \ge t \,\,\, and \,\,\,x^3 + y^3 + z^3 \le t.$$
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Tags: algebra, inequalities
rms
21.08.2024 10:06
parmenides51 wrote: Find all real numbers $x, y,z,t \in [0, \infty)$ so that $$x + y + z \le t, \,\,\, x^2 + y^2 + z^2 \ge t \,\,\, and \,\,\,x^3 + y^3 + z^3 \le t.$$
By adding the inequalities term by term \( x + y + z \leq t \), \( -2x^2 - 2y^2 - 2z^2 \leq -2t \), and \( x^3 + y^3 + z^3 \leq t \), we obtain:
\[
x(1-x)^2 + y(1-y)^2 + z(1-z)^2 \leq 0
\]
Since \( x, y, z \in [0, \infty) \), it follows that \( x, y, z \in \{0, 1\} \).
- If \( x = y = z = 0 \), then \( t = 0 \).
- If exactly two of the numbers \( x, y, z \) are zero, then \( 1 \leq t \geq 1 \), hence \( t = 1 \) and \( (x, y, z) \in \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\} \).
- If exactly one of the numbers \( x, y, z \) is zero, then \( 2 \leq t \geq 2 \), so \( t = 2 \) and \( (x, y, z) \in \{(1, 1, 0), (1, 0, 1), (0, 1, 1)\} \).
- If \( x = y = z = 1 \), we deduce that \( t = 3 \).