The discussion revolves around proving the inequality \(3!\sqrt{x}+2!y+1!z^2\le 13\) under the condition that \(x^2+y^2+z^2+xyz=4\) with \(x, y, z \ge 0\). The scope includes mathematical reasoning and attempts to clarify the expressions involved.
There is no consensus on the understanding of the initial line or the correctness of the expressions, as confusion and questions remain among participants.
Participants have noted potential typographical errors and expressed uncertainty about the initial conditions and expressions, which may affect the clarity of the discussion.
anemone said:If $x^2+y^2+{\color{red}z}^2+xyz=4$ and that $x,\,y,\,x\ge 0$, prove $3!\sqrt{x}+2!y+1!z^2\le 13$.
Olinguito said:Is there a typo? Should it be …
Olinguito said:My attempt.
We have (by AM–GM or otherwise):
$$\begin{array}{rcl}6\sqrt x &\le& x^2+\dfrac{81}{16}+1+1 \\ 2y &\le& y^2+1 \\ z^2 &\le& z^2+xyz\end{array}$$
$\begin{array}{rl}\implies & 6\sqrt x+2y+z^2 \\ \le & x^2+y^2+z^2+xyz+\dfrac{129}{16}\ =\ \dfrac{193}{16}\ <\ 13.\end{array}$
Olinguito said:$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)