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

[anemone]

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

 

[Olinguito]

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]

Is there a typo? Should it be …

Indeed it is a typo(Tmi), sorry about that, and I just fixed it! Thanks for catching the typo!(Yes)

 

[Olinguito]

My attempt.

Spoiler

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

 

[kaliprasad]

My attempt.

Spoiler

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]

Spoiler

$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)

 

[kaliprasad]

Spoiler

$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)

Spoiler

I never said it was wrong. I said In did not understand. Thanks or clarifying. It is correct