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- Feb 14, 2012

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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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- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,599

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

Last edited:

- Apr 22, 2018

- 251

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

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- #3

- Feb 14, 2012

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Indeed it is a typo, sorry about that, and I just fixed it! Thanks for catching the typo!Is there a typo? Should it be …

- Apr 22, 2018

- 251

$$\begin{array}{rcl}6\sqrt x &\le& x^2+\dfrac{81}{16}+1+1 \\ 2y &\le& y^2+1 \\ z^2 &\le& z^2+xyz\end{array}$$

$\begin{array}{rl}\implies & 6\sqrt x+2y+z^2 \\ \le & x^2+y^2+z^2+xyz+\dfrac{129}{16}\ =\ \dfrac{193}{16}\ <\ 13.\end{array}$

- Mar 31, 2013

- 1,283

I do not understand the 1st line

$$\begin{array}{rcl}6\sqrt x &\le& x^2+\dfrac{81}{16}+1+1 \\ 2y &\le& y^2+1 \\ z^2 &\le& z^2+xyz\end{array}$$

$\begin{array}{rl}\implies & 6\sqrt x+2y+z^2 \\ \le & x^2+y^2+z^2+xyz+\dfrac{129}{16}\ =\ \dfrac{193}{16}\ <\ 13.\end{array}$

- Apr 22, 2018

- 251

(AM–GM). Did I get it wrong?

- Mar 31, 2013

- 1,283

(AM–GM). Did I get it wrong?

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