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

AI Thread Summary
The discussion revolves around proving the inequality 3!√{x} + 2!y + 1!z² ≤ 13 under the condition x² + y² + z² + xyz = 4 with x, y, z ≥ 0. Participants clarify a typo in the original expression, confirming that it should read z² instead of an incorrect variable. There is a request for further clarification on the first line of the problem statement. The conversation highlights the importance of accurate notation in mathematical proofs. Overall, the focus remains on resolving the inequality and ensuring the conditions are correctly stated.
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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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Is there a typo? Should it be …

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 …


Indeed it is a typo(Tmi), sorry about that, and I just fixed it! Thanks for catching the typo!(Yes)
 
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:
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
 
$$\frac{x^2+\dfrac{81}{16}+1+1}4\ \ge\ \sqrt[4]{(x^2)\left(\frac{81}{16}\right)(1)(1)}$$
(AM–GM). Did I get it wrong? (Sweating)
 
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)


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

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