<s>[INST] Show $2(x+y+z)-xyz\leq 10$ if $x^2+y^2+z^2=9$ If $x,y,z$ are real and $x^2+y^2+z^2=9$, how can we prove that $2(x+y+z)-xyz\leq 10$?
Please provide a solution without the use of calculus. I know the solution in that way.
 [/INST] without loss of generality we can assume that $|x|\le |y|\le |z|$,and  Note that $3z^2\ge x^2+y^2+z^2$, which implies that $z^2\ge 3$
we have
\begin{align}
[2(x+y+z)-xyz]^2&=[2(x+y)+z(2-xy)]^2\\
&\le [(x+y)^2+z^2][4+(2-xy)^2]\\
&=(9+2xy)(8-4xy+(xy)^2))\\
&=2(xy)^3+(xy)^2-20(xy)+72\\
&=(xy+2)^2(2xy-7)+100
\end{align}
from $z^2\ge 3$ it follows that $2xy\le x^2+y^2=9-z^2\le 6$,then
$$[2(x+y+z)-xyz]^2\le 100$$
or equivalently 
$$2(x+y+z)-xyz\le 10$$
</s>