If $x, y, z > 0$ and $x y z = 1$ .
Then

x^{2} + y^{2} + z^{2} \leq x^{3} + y^{3} + z^{3}

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x^{2} + y^{2} + z^{2} \geq x^{3} + y^{3} + z^{3}

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(x^{4} + y^{4} + z^{4}) \geq (x^{3} + y^{3} + z^{3})

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(x^{4} + y^{4} + z^{4}) \leq (x^{3} + y^{3} + z^{3})

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Solution

The correct options are
A $(x^{4} + y^{4} + z^{4}) \geq (x^{3} + y^{3} + z^{3})$
B $x^{2} + y^{2} + z^{2} \leq x^{3} + y^{3} + z^{3}$
Consider , $f (u) = x^{u} + y^{u} + z^{u}$
$f^{'} (u) = x^{u} l n (x) + y^{u} l n (y) + z^{u} l n (z)$
$f^{'} (0) = x^{0} l n (x) + y^{0} l n (y) + z^{0} l n (z)$
$= l n (x y z) = - l n (1) = 0$
So $f^{'} (0) = 0 . . . . . (i)$
$f^{''} (u) = x^{u} (l n (x))^{2} + y^{u} (l n (y))^{2} + z^{u} (l n (z))^{2}$
Thus, $f^{''} (u) > 0..... (i i)$
So $f^{'} (u)$ is increasing $f n c$ .
So $f^{'} (u) \geq f^{'} (0)$ for $u \geq 0$
Thus, $f (u)$ is itself increasing for $u \geq 0$
Thus, $x^{2} + y^{2} + z^{2} \leq x^{3} + y^{3} + z^{3}$
Option (A).

Also, $(x^{4} + y^{4} + z^{4}) \geq (x^{3} + y^{3} + z^{3})$ .
So, option (C), as well.

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