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

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,599
If $x^2+y^2+z^2+xyz=4$ and that $x,\,y,\,x\ge 0$, prove $3!\sqrt{x}+2!y+1!z^2\le 13$.
 
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Olinguito

Well-known member
Apr 22, 2018
251
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 $3!\sqrt{x}+2!y+1!z^2\le 13$.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Olinguito

Well-known member
Apr 22, 2018
251
My attempt.

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

kaliprasad

Well-known member
Mar 31, 2013
1,283
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
Apr 22, 2018
251
$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)

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