# [SOLVED]Inequality involves radical, square and factorial expression 3!√{x}+2!y+1!z^2⩽ 13

#### anemone

##### MHB POTW Director
Staff member
If $x^2+y^2+z^2+xyz=4$ and that $x,\,y,\,x\ge 0$, prove $3!\sqrt{x}+2!y+1!z^2\le 13$.

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#### Olinguito

##### Well-known member
Is there a typo? Should it be …

If $x^2+y^2+{\color{red}z}^2+xyz=4$ and that $x,\,y,\,x\ge 0$, prove $3!\sqrt{x}+2!y+1!z^2\le 13$.

#### anemone

##### MHB POTW Director
Staff member
Is there a typo? Should it be …
Indeed it is a typo , sorry about that, and I just fixed it! Thanks for catching the typo! #### Olinguito

##### Well-known member
My attempt.

We have (by AM–GM or otherwise):
$$\begin{array}{rcl}6\sqrt x &\le& x^2+\dfrac{81}{16}+1+1 \\ 2y &\le& y^2+1 \\ z^2 &\le& z^2+xyz\end{array}$$
$\begin{array}{rl}\implies & 6\sqrt x+2y+z^2 \\ \le & x^2+y^2+z^2+xyz+\dfrac{129}{16}\ =\ \dfrac{193}{16}\ <\ 13.\end{array}$

##### Well-known member
My attempt.

We have (by AM–GM or otherwise):
$$\begin{array}{rcl}6\sqrt x &\le& x^2+\dfrac{81}{16}+1+1 \\ 2y &\le& y^2+1 \\ z^2 &\le& z^2+xyz\end{array}$$
$\begin{array}{rl}\implies & 6\sqrt x+2y+z^2 \\ \le & x^2+y^2+z^2+xyz+\dfrac{129}{16}\ =\ \dfrac{193}{16}\ <\ 13.\end{array}$
I do not understand the 1st line

#### Olinguito

##### Well-known member
$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? $$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? 